This sheet lists the Structure of Atom Class 11 formulas from the NCERT Chemistry Part I textbook, Chapter 2. It covers the charge and mass of the electron, the relations of electromagnetic radiation, Planck’s quantum formula, the photoelectric equation, the Rydberg and Bohr formulas for the hydrogen atom, the de Broglie relation, Heisenberg’s uncertainty principle, and the quantum-number counting rules.
Each formula is grouped by topic, with a symbol table that gives the meaning and unit of every quantity, guidance on when each formula is used, worked examples with original numbers, and chapter-specific mistakes to avoid. For the detailed explanations and derivations, browse the Class 11 Chemistry formulas collection.
Structure of Atom Class 11 Formulas at a Glance
This table indexes every formula on the page. Find the row you need, then read the grouped list below for its conditions.
| Purpose (what you are finding) | Formula |
|---|---|
| Charge-to-mass ratio of an electron | \( \frac{e}{m_e} = 1.758820 \times 10^{11}\ \text{C kg}^{-1} \) |
| Mass of an electron (from the measured charge and charge-to-mass ratio) | \( m_e = \frac{e}{(e/m_e)} = 9.1094 \times 10^{-31}\ \text{kg} \) |
| Charge on a body is a whole-number multiple of \(e\) (Millikan’s oil drop) | \( q = ne,\ n = 1, 2, 3, \dots \) |
| Atomic number of a neutral atom | \( Z = \text{number of protons} = \text{number of electrons} \) |
| Mass number (total nucleons) | \( A = Z + n \) |
| Symbol of a nuclide (derived: neutrons \(= A – Z\)) | \( {}_Z^A X \) |
| Speed, frequency and wavelength of a wave | \( c = \nu\lambda \) |
| Wavenumber from wavelength | \( \bar{\nu} = \frac{1}{\lambda} \) |
| Energy of one quantum of radiation | \( E = h\nu \) |
| Energy of a photon when the wavelength is given (from \(c=\nu\lambda\) and \(E=h\nu\)) | \( E = \frac{hc}{\lambda} \) |
| Allowed energies of a quantised oscillator | \( E = nh\nu,\ n = 1, 2, 3, \dots \) |
| Einstein’s photoelectric equation | \( h\nu = h\nu_0 + \frac{1}{2}m_e v^2 \) |
| Kinetic energy of the ejected electron (rearranged photoelectric equation; valid above the threshold frequency) | \( \frac{1}{2}m_e v^2 = h(\nu – \nu_0) \) |
| Wavenumber of a hydrogen spectral line (Rydberg formula) | \( \bar{\nu} = 109{,}677\left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)\ \text{cm}^{-1} \) |
| Frequency of radiation emitted or absorbed between two states (Bohr’s frequency rule) | \( \nu = \frac{E_2 – E_1}{h} = \frac{\Delta E}{h} \) |
| Quantisation of the electron’s angular momentum | \( m_e v r = n\frac{h}{2\pi} \) |
| Radius of the \(n\)th orbit of hydrogen or a hydrogen-like ion | \( r_n = \frac{52.9\ n^2}{Z}\ \text{pm} \) |
| Energy of the \(n\)th stationary state of hydrogen or a hydrogen-like ion | \( E_n = -\frac{2.18 \times 10^{-18}\ Z^2}{n^2}\ \text{J} \) |
| Energy absorbed or emitted in a transition | \( \Delta E = 2.18 \times 10^{-18}\left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)\ \text{J} \) |
| Frequency of the photon in a transition (derived from \(\Delta E/h\)) | \( \nu = 3.29 \times 10^{15}\left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)\ \text{Hz} \) |
| Wavelength of a photon from its energy (from \(E = hc/\lambda\)) | \( \lambda = \frac{hc}{E} \) |
| de Broglie wavelength of a moving particle | \( \lambda = \frac{h}{mv} = \frac{h}{p} \) |
| Heisenberg uncertainty in position and momentum | \( \Delta x \cdot \Delta p_x \geq \frac{h}{4\pi} \) |
| Heisenberg uncertainty in position and velocity (derived from \(\Delta p_x = m\,\Delta v_x\)) | \( \Delta x \cdot \Delta v_x \geq \frac{h}{4\pi m} \) |
| Number of orbitals in a shell | \( n^2 \) |
| Number of orbitals in a subshell | \( 2l + 1 \) |
| Nodes in an orbital (total, angular, radial) | \( n – 1 = l + (n – l – 1) \) |
| Schrödinger equation (wave-mechanical statement of the model) | \( \hat{H}\Psi = E\Psi \) |
All Formulas, Grouped by Topic
Formulas appear under the chapter’s own sub-topics, in the order the textbook presents them. Page references are to the NCERT Class 11 Chemistry Part I textbook (Rationalised NCERT), available as an official PDF at ncert.nic.in.
Sub-atomic Particles and Nuclear Notation
Charge-to-mass ratio of the electron. Thomson balanced electric and magnetic deflections of a cathode-ray beam to measure this ratio (NCERT, p. 32):
\[ \frac{e}{m_e} = 1.758820 \times 10^{11}\ \text{C kg}^{-1} \]
Mass of the electron. Millikan’s oil-drop charge combined with Thomson’s ratio gives (NCERT, p. 32):
\[ m_e = \frac{e}{e/m_e} = 9.1094 \times 10^{-31}\ \text{kg} \]
Quantisation of charge. Every charge Millikan measured was an integral multiple of \(e\), that is \( q = ne \) with \( n = 1, 2, 3, \dots \) (NCERT, p. 33).
Atomic number and mass number. Protons give the nucleus its charge; protons plus neutrons give it its mass (NCERT, p. 36):
\[ Z = \text{protons} = \text{electrons (neutral atom)},\qquad A = Z + n \]
Neutrons are therefore \( A – Z \), for a neutral atom or an ion alike. The nuclide is written with the mass number above and the atomic number below: \( {}_Z^A X \).
Electromagnetic Radiation and Planck’s Quantum Theory
Speed, frequency and wavelength. In vacuum all electromagnetic radiation travels at \( c = 3.0 \times 10^8\ \text{m s}^{-1} \) (NCERT, pp. 38-39):
\[ c = \nu\lambda \]
Wavenumber. Spectroscopy usually works with wavelengths per unit length (NCERT, p. 39):
\[ \bar{\nu} = \frac{1}{\lambda} \]
Energy of one quantum. Planck assumed radiation is emitted or absorbed only in discrete chunks, with energy proportional to frequency (NCERT, pp. 40-41):
\[ E = h\nu \]
That is why the allowed values are the staircase-like set \( E = 0, h\nu, 2h\nu, 3h\nu, \dots, nh\nu \): an oscillator can sit on a step, never between steps.
Photon energy from wavelength. Substituting \( \nu = c/\lambda \) into \( E = h\nu \) gives the form used in most numericals, and its inverse (NCERT, p. 44):
\[ E = \frac{hc}{\lambda},\qquad \lambda = \frac{hc}{E} \]
Photoelectric Effect
Einstein’s photoelectric equation. A photon gives its energy \( h\nu \) to an electron; part pays the work function \( W_0 = h\nu_0 \) and the rest becomes kinetic energy (NCERT, pp. 42-43):
\[ h\nu = h\nu_0 + \frac{1}{2}m_e v^2 \]
Kinetic energy of the ejected electron. Rearranged, this is the form to use when the question gives two frequencies (NCERT, p. 44):
\[ \frac{1}{2}m_e v^2 = h(\nu – \nu_0) \]
It holds only when \( \nu \geq \nu_0 \). Below the threshold frequency no electrons are ejected, however bright the light — the chapter’s potassium example makes this concrete (NCERT, p. 42).
Atomic Spectra of Hydrogen
Rydberg formula. Balmer found the visible lines first, then Rydberg generalised the pattern to every series of the hydrogen spectrum (NCERT, p. 45):
\[ \bar{\nu} = 109{,}677\left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)\ \text{cm}^{-1},\qquad n_2 \gt n_1 \]
The constant \( 109{,}677\ \text{cm}^{-1} \) is the Rydberg constant for hydrogen written as a wavenumber; the same chapter reports it as \( R_H = 2.18 \times 10^{-18}\ \text{J} \) when energies are involved (NCERT, p. 47).
The series are named after their lower level \( n_1 \) (NCERT, p. 46):
| Series | \( n_1 \) | \( n_2 \) | Spectral region |
|---|---|---|---|
| Lyman | 1 | 2, 3, 4, … | Ultraviolet |
| Balmer | 2 | 3, 4, 5, … | Visible |
| Paschen | 3 | 4, 5, 6, … | Infrared |
| Brackett | 4 | 5, 6, 7, … | Infrared |
| Pfund | 5 | 6, 7, 8, … | Infrared |
Bohr’s Model for Hydrogen Atom
Bohr’s frequency rule. Radiation appears only when the electron jumps between two stationary states (NCERT, p. 47):
\[ \nu = \frac{E_2 – E_1}{h} = \frac{\Delta E}{h} \]
Quantisation of angular momentum. Only orbits whose angular momentum is a whole multiple of \( h/2\pi \) are allowed — this is the postulate that stops the electron spiralling in (NCERT, p. 47):
\[ m_e v r = n\frac{h}{2\pi},\qquad n = 1, 2, 3, \dots \]
Radius of the \(n\)th orbit. For hydrogen \( r_n = n^2 a_0 \) with the Bohr radius \( a_0 = 52.9\ \text{pm} \); for hydrogen-like ions such as He⁺, Li²⁺ and Be³⁺ the radius shrinks with nuclear charge (NCERT, pp. 47-48):
\[ r_n = \frac{52.9\ n^2}{Z}\ \text{pm} \]
Energy of a stationary state. The negative sign means the electron is bound: its energy is lower than that of a free electron at rest, which is assigned zero (NCERT, pp. 47-48):
\[ E_n = -\frac{2.18 \times 10^{-18}\ Z^2}{n^2}\ \text{J} \]
Energy of a transition. The difference of two \( E_n \) values gives (NCERT, p. 49):
\[ \Delta E = 2.18 \times 10^{-18}\left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)\ \text{J} \]
For emission \( n_i \gt n_f \), so \( \Delta E \) is negative — energy is released. For absorption \( n_f \gt n_i \) and \( \Delta E \) is positive.
Frequency and wavenumber of a spectral line. Divide \( \Delta E \) by \( h \) or by \( hc \) for the two shortcut forms (NCERT, p. 49):
\[ \nu = 3.29 \times 10^{15}\left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)\ \text{Hz} \]
\[ \bar{\nu} = 1.09677 \times 10^{7}\left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)\ \text{m}^{-1} \]
de Broglie Relation and Heisenberg Uncertainty Principle
de Broglie relation. Matter, like radiation, has a wavelength; the electron’s wave nature was later confirmed by electron diffraction (NCERT, p. 50):
\[ \lambda = \frac{h}{mv} = \frac{h}{p} \]
Heisenberg uncertainty principle. The exact position and exact momentum of an electron cannot be known simultaneously; the product of their uncertainties has a floor (NCERT, p. 51):
\[ \Delta x \cdot \Delta p_x \geq \frac{h}{4\pi} \]
Dividing by \( m \) gives the velocity form used in numericals:
\[ \Delta x \cdot \Delta v_x \geq \frac{h}{4\pi m} \]
The product \( \Delta x\,\Delta v \) equals \( h/4\pi m \), so a large mass makes the uncertainty negligible — about \( 10^{-28}\ \text{m}^2\ \text{s}^{-1} \) for a milligram object, but roughly \( 10^{-4}\ \text{m}^2\ \text{s}^{-1} \) for an electron (NCERT, p. 52). That is why Bohr orbits cannot exist for the electron.
Quantum Numbers and Orbitals
Four quantum numbers label an electron, and the counting rules follow directly from their allowed values (NCERT, pp. 55-56):
- Principal \(n\): positive integer \( 1, 2, 3, \dots \); fixes shell, size and largely energy. Number of orbitals in a shell: \( n^2 \).
- Azimuthal \(l\): \( 0, 1, 2, \dots, n-1 \); fixes sub-shell shape (s, p, d, f). Number of sub-shells in a shell: \( n \).
- Magnetic \(m_l\): \( -l \) to \( +l \), giving \( 2l + 1 \) orbitals per sub-shell (one s, three p, five d).
- Spin \(m_s\): \( +\frac{1}{2} \) or \( -\frac{1}{2} \) (NCERT, p. 56).
Number of nodes in an orbital (NCERT, pp. 58-59):
\[ \text{total nodes} = n – 1 = \underbrace{l}_{\text{angular}} + \underbrace{(n – l – 1)}_{\text{radial}} \]
The model itself rests on the Schrödinger equation (NCERT, p. 54):
\[ \hat{H}\Psi = E\Psi \]
where \( \Psi \) is the electron’s wave function; \( |\Psi|^2 \) gives the probability density of finding the electron at a point.
What Each Symbol Means
The table lists every symbol used on this page. Units follow the SI conventions the chapter uses.
| Symbol | What it means | Unit |
|---|---|---|
| \( e \) | Magnitude of the charge on an electron | C |
| \( m_e \) | Mass of an electron | kg |
| \( q \) | Electric charge on a body (Millikan’s drops) | C |
| \( Z \) | Atomic number: protons in the nucleus; equals electrons in a neutral atom | count (dimensionless) |
| \( A \) | Mass number: total protons + neutrons | count (dimensionless) |
| \( n \) (in \( A = Z + n \)) | Number of neutrons | count (dimensionless) |
| \( \nu \) | Frequency of electromagnetic radiation | Hz (s⁻¹) |
| \( \lambda \) | Wavelength | m (1 nm = \( 10^{-9} \) m; 1 Å = \( 10^{-10} \) m) |
| \( c \) | Speed of light in vacuum, \( 3.0 \times 10^8 \) | m s⁻¹ |
| \( \bar{\nu} \) | Wavenumber: number of wavelengths per unit length | m⁻¹ (often cm⁻¹) |
| \( h \) | Planck’s constant, \( 6.626 \times 10^{-34} \) | J s |
| \( \nu_0 \) | Threshold frequency of a metal | Hz |
| \( W_0 \) | Work function, \( h\nu_0 \): minimum energy to eject an electron | J (chapter table also uses eV) |
| \( v \) | Speed of a particle (electron, photoelectron) | m s⁻¹ |
| \( E \) | Energy of a photon or of a stationary state | J |
| \( \Delta E \) | Energy difference between two states | J |
| \( R_H \) | Rydberg constant (energy form), \( 2.18 \times 10^{-18} \) | J |
| \( r_n \) | Radius of the \(n\)th orbit | pm |
| \( a_0 \) | Bohr radius: radius of the first orbit of hydrogen, \( 52.9 \) | pm |
| \( p \) | Momentum of a particle | kg m s⁻¹ |
| \( \Delta x \) | Uncertainty in position | m |
| \( \Delta p_x \) | Uncertainty in momentum | kg m s⁻¹ |
| \( \Delta v_x \) | Uncertainty in velocity | m s⁻¹ |
| \( n \) (in Bohr formulas) | Principal quantum number, \( 1, 2, 3, \dots \) | integer |
| \( n_i, n_f \) | Initial and final principal quantum numbers of a transition | integer |
| \( n_1, n_2 \) | Lower and upper levels in the Rydberg formula, with \( n_2 \gt n_1 \) | integer |
| \( l \) | Azimuthal (shape) quantum number, \( 0 \) to \( n – 1 \) | dimensionless |
| \( m_l \) | Magnetic orbital quantum number, \( -l \) to \( +l \) | dimensionless |
| \( m_s \) | Electron spin quantum number | \( \pm \frac{1}{2} \) (dimensionless) |
| \( \Psi \) | Wave function of an electron in an atom (an atomic orbital) | — |
| \( |\Psi|^2 \) | Probability density at a point | per unit volume |
| \( \hat{H} \) | Hamiltonian operator in the Schrödinger equation | — |
Two reading notes. First, \( n \) does triple duty on this page: number of neutrons in \( A = Z + n \), the principal quantum number in Bohr’s formulas, and the integer counter in \( q = ne \) — always check the context. Second, do not confuse three similar symbols: \( \nu \) (frequency, Hz), \( v \) (velocity, m s⁻¹) and \( \bar{\nu} \) (wavenumber, m⁻¹ or cm⁻¹).
When to Use Each Formula
This table is the decision aid: match your question to the situation, then apply the formula in the right units.
| When you need… | Reach for |
|---|---|
| Wavelength ↔ frequency of any electromagnetic wave | \( c = \nu\lambda \), with \( c = 3.0 \times 10^8\ \text{m s}^{-1} \) |
| Number of wavelengths per unit length (spectroscopy) | \( \bar{\nu} = \frac{1}{\lambda} \) |
| Energy of one photon when frequency is given | \( E = h\nu \) |
| Energy of one photon when wavelength is given (most numericals) | \( E = \frac{hc}{\lambda} \) |
| Wavelength of a photon from its energy | \( \lambda = \frac{hc}{E} \) |
| Kinetic energy of photoelectrons (only if \( \nu \geq \nu_0 \)) | \( \frac{1}{2}m_e v^2 = h(\nu – \nu_0) \) |
| Wavenumber of a hydrogen spectral line (keep \( n_2 \gt n_1 \)) | \( \bar{\nu} = 109{,}677\left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)\ \text{cm}^{-1} \) |
| Identifying the series of a transition | lower level \( n_1 \): 1 = Lyman (UV), 2 = Balmer (visible), 3 = Paschen, 4 = Brackett, 5 = Pfund (IR) |
| Energy of a stationary state of H, He⁺, Li²⁺, Be³⁺ | \( E_n = -\frac{2.18 \times 10^{-18} Z^2}{n^2}\ \text{J} \) |
| Radius of an orbit of a hydrogen-like species | \( r_n = \frac{52.9\ n^2}{Z}\ \text{pm} \) |
| Energy absorbed or released in a transition | \( \Delta E = 2.18 \times 10^{-18}\left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)\ \text{J} \) |
| Frequency of the emitted or absorbed photon | \( \nu = 3.29 \times 10^{15}\left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)\ \text{Hz} \) |
| Wavelength of any moving particle (electron, ball) | \( \lambda = \frac{h}{mv} \) — mass must be in kg |
| Uncertainty in position, momentum or velocity of an electron | \( \Delta x \cdot \Delta p_x \geq \frac{h}{4\pi} \) or \( \Delta x \cdot \Delta v_x \geq \frac{h}{4\pi m} \) |
| Protons, neutrons and electrons of a species | \( Z = p = e \) (neutral atom); \( A = Z + n \) |
| Counting orbitals or nodes | \( n^2 \) per shell; \( 2l + 1 \) per sub-shell; total nodes \( n – 1 \) |
Worked Examples
Three examples with original numbers show the common patterns: applying the photon formulas directly, working backwards through the photoelectric equation, and using the alternative transition form. For more revision sheets from other chapters, visit the Chemistry formulas index.
Example 1: Frequency and energy of a photon of green light (wavelength 550 nm)
- Step 1: Convert wavelength to metres: \( 550\ \text{nm} = 550 \times 10^{-9}\ \text{m} \).
- Step 2: Use \( c = \nu\lambda \) to find the frequency.
\[ \nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8\ \text{m s}^{-1}}{550 \times 10^{-9}\ \text{m}} = 5.45 \times 10^{14}\ \text{Hz} \]
Step 3: Use \( E = h\nu \) with \( h = 6.626 \times 10^{-34}\ \text{J s} \).
\[ E = (6.626 \times 10^{-34}\ \text{J s})(5.45 \times 10^{14}\ \text{s}^{-1}) = 3.61 \times 10^{-19}\ \text{J} \]
Final answer: \( \nu = 5.45 \times 10^{14}\ \text{Hz} \) and \( E = 3.61 \times 10^{-19}\ \text{J} \). The same result comes directly from \( E = hc/\lambda \).
Example 2: Work function and threshold frequency from the photoelectric equation
Step 1: Find the energy of one photon of the \( 400\ \text{nm} \) light.
\[ E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}\ \text{J s})(3.0 \times 10^8\ \text{m s}^{-1})}{400 \times 10^{-9}\ \text{m}} = 4.97 \times 10^{-19}\ \text{J} \]
Step 2: In \( h\nu = W_0 + \tfrac{1}{2}m_e v^2 \), the photon energy pays the work function plus the electron’s kinetic energy \( 0.83 \times 10^{-19}\ \text{J} \).
\[ W_0 = E – \text{KE} = 4.97 \times 10^{-19} – 0.83 \times 10^{-19} = 4.14 \times 10^{-19}\ \text{J} \]
Step 3: Convert the work function to a threshold frequency with \( W_0 = h\nu_0 \).
\[ \nu_0 = \frac{W_0}{h} = \frac{4.14 \times 10^{-19}\ \text{J}}{6.626 \times 10^{-34}\ \text{J s}} = 6.25 \times 10^{14}\ \text{Hz} \]
Final answer: \( W_0 = 4.14 \times 10^{-19}\ \text{J} \) and \( \nu_0 = 6.25 \times 10^{14}\ \text{Hz} \). Since \( 400\ \text{nm} \) light has frequency \( 7.5 \times 10^{14}\ \text{Hz} \), which exceeds \( \nu_0 \), electrons are indeed ejected.
Example 3: Wavelength of the photon emitted in a hydrogen transition (n = 4 to n = 2)
Step 1: This is emission, so \( n_i = 4 \), \( n_f = 2 \).
Use the transition energy formula.
\[ \Delta E = 2.18 \times 10^{-18}\left(\frac{1}{4^2} – \frac{1}{2^2}\right)\ \text{J} = 2.18 \times 10^{-18}\left(\frac{1}{16} – \frac{1}{4}\right)\ \text{J} = -4.09 \times 10^{-19}\ \text{J} \]
Step 2: The negative sign means energy is released.
The photon carries the magnitude \( |\Delta E| = 4.09 \times 10^{-19}\ \text{J} \).
Step 3: Find the wavelength from \( E = hc/\lambda \).
\[ \lambda = \frac{hc}{|\Delta E|} = \frac{(6.626 \times 10^{-34}\ \text{J s})(3.0 \times 10^8\ \text{m s}^{-1})}{4.09 \times 10^{-19}\ \text{J}} = 4.86 \times 10^{-7}\ \text{m} = 486\ \text{nm} \]
Final answer: \( \lambda = 486\ \text{nm} \). The transition ends at \( n_f = 2 \), so the line belongs to the Balmer series in the visible region.
Common Mistakes to Avoid
These are the errors this chapter actually produces in tests — wrong symbol, wrong sign, a forgotten square, or a wrong unit.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Writing \( \nu \), \( v \) and \( \bar{\nu} \) interchangeably | \( \nu \) is frequency in Hz (s⁻¹), \( v \) is velocity in m s⁻¹, \( \bar{\nu} \) is wavenumber in m⁻¹ or cm⁻¹ | Read the unit after the number: Hz, m s⁻¹ and m⁻¹ name three different quantities |
| Putting \( n_2 \lt n_1 \) in the Rydberg formula and getting a negative wavenumber | Always \( n_2 \gt n_1 \): the bracket \( \left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right) \) must be positive | A wavenumber or frequency can never come out negative — swap the levels if it does |
| Reporting the energy of an emitted line as a positive \( \Delta E \) with the sign ignored | Emission has \( n_i \gt n_f \), so \( \Delta E \lt 0 \); the photon takes \( |\Delta E| = h\nu \) | Emission (electron falls) → photon energy is the magnitude of \( \Delta E \); absorption → \( \Delta E \) positive |
| Forgetting to square \( Z \) in \( E_n = -\frac{2.18 \times 10^{-18} Z^2}{n^2}\ \text{J} \) for hydrogen-like ions | Energy scales with \( Z^2 \), radius scales with \( 1/Z \) | He⁺ ground state: \( 2^2 = 4 \) times the hydrogen value, \( -8.72 \times 10^{-18}\ \text{J} \) |
| Using mass in grams in the de Broglie or Heisenberg formulas | Mass must be in kg because \( 1\ \text{J} = 1\ \text{kg m}^2\ \text{s}^{-2} \) | Do a unit check: \( h/mv \) must end in metres — grams will not cancel to metres |
Frequently Asked Questions
Why is the electron’s energy negative in the Bohr energy formula?
Zero energy is assigned to a free electron at rest, infinitely far from the nucleus (\( n = \infty \)). An electron bound in the atom has less energy than that reference state, so every \( E_n \) is negative. The most negative value (\( n = 1 \)) is the ground state — the most stable orbit. As \( n \) increases, \( E_n \) becomes less negative and approaches zero.
How do I identify the spectral series of a hydrogen line?
Look at the lower level of the transition, \( n_1 \): Lyman (\( n_1 = 1 \), ultraviolet), Balmer (\( n_1 = 2 \), visible), Paschen (\( n_1 = 3 \), infrared), Brackett (\( n_1 = 4 \), infrared), Pfund (\( n_1 = 5 \), infrared). For example, a \( 4 \rightarrow 2 \) transition ends at \( n_1 = 2 \), so it is a Balmer line in the visible region, and the example above gives \( 486\ \text{nm} \).
What is the difference between an orbit and an orbital?
A Bohr orbit is a definite circular path — a concept the Heisenberg uncertainty principle rules out, because the electron’s exact position and velocity cannot both be known. An orbital is a quantum-mechanical region, described by the wave function \( \Psi \), that encloses about 90% probability of finding the electron; \( |\Psi|^2 \) at a point is the probability density.
Orbits have no real existence; orbitals are the working picture of the quantum model.
How do I find the number of photons emitted per second by a source?
Divide the power of the source by the energy of one photon: photons per second \( = \dfrac{P}{hc/\lambda} \). Example: a \( 5.0\ \text{mW} \) laser at \( 633\ \text{nm} \) has photon energy \( E = hc/\lambda = 3.14 \times 10^{-19}\ \text{J} \), so it emits \( \dfrac{5.0 \times 10^{-3}}{3.14 \times 10^{-19}} \approx 1.6 \times 10^{16} \) photons each second.
Reference: NCERT Class 11 Chemistry textbook, chapter Structure of Atom.
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