Modern physics formula sheet — photoelectric, Bohr, nuclear, semiconductors

Question

Light of wavelength 200 nm falls on a metal surface with work function 4.2 eV. Find (a) the maximum kinetic energy of photoelectrons, (b) the stopping potential, and (c) the threshold wavelength.

(JEE Main / NEET — Dual Nature of Radiation and Matter)

Modern Physics Topic Map

flowchart TD
    A["Modern Physics"] --> B["Photoelectric Effect"]
    A --> C["Bohr's Atom"]
    A --> D["Nuclear Physics"]
    A --> E["Semiconductors"]
    B --> B1["KE_max = hf - phi"]
    B --> B2["eV_0 = KE_max"]
    C --> C1["E_n = -13.6/n² eV"]
    C --> C2["r_n = 0.529 n² Angstrom"]
    D --> D1["E = mc²"]
    D --> D2["Mass defect and binding energy"]
    E --> E1["p-n junction, diode, LED"]
    E --> E2["Forward/Reverse bias"]

Solution — Step by Step

Energy of a photon: $E = \dfrac{hc}{\lambda}$

$h = 6.63 \times 10^{-34}$ J $\cdot$ s, $c = 3 \times 10^8$ m/s, $\lambda = 200$ nm = $200 \times 10^{-9}$ m

E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{200 \times 10^{-9}} = \frac{19.89 \times 10^{-26}}{2 \times 10^{-7}} = 9.945 \times 10^{-19} \text{ J}

Converting to eV: $E = \dfrac{9.945 \times 10^{-19}}{1.6 \times 10^{-19}} = 6.22$ eV

Shortcut: $E(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})} = \dfrac{1240}{200} = 6.2$ eV

Einstein’s photoelectric equation: $KE_{\max} = h\nu - \phi$

KE_{\max} = 6.2 - 4.2 = \mathbf{2.0 \text{ eV}}

Stopping potential $V_0$ is defined by: $eV_0 = KE_{\max}$

\boxed{V_0 = \frac{KE_{\max}}{e} = 2.0 \text{ V}}

At threshold, $KE_{\max} = 0$ , so all photon energy equals the work function:

\phi = \frac{hc}{\lambda_0} \implies \lambda_0 = \frac{hc}{\phi} = \frac{1240}{4.2} = 295.2 \text{ nm}

\boxed{\lambda_0 \approx 295 \text{ nm}}

Any wavelength longer than 295 nm (lower energy) will not eject electrons from this metal.

Why This Works

Einstein’s photoelectric equation treats light as particles (photons), each carrying energy $h\nu$ . A photon transfers all its energy to one electron. Part of that energy ( $\phi$ , the work function) is used to free the electron from the metal. The rest becomes kinetic energy. This particle model of light explained observations that wave theory could not: the instantaneous emission and the frequency threshold.

Alternative Method — Key Formulas Reference

Bohr Model (Hydrogen):

Energy: $E_n = -\dfrac{13.6}{n^2}$ eV
Radius: $r_n = 0.529 \times n^2$ angstrom
Wavelength of emitted photon: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ where $R = 1.097 \times 10^7$ m $^{-1}$

Nuclear Physics:

Mass defect: $\Delta m = Zm_p + (A-Z)m_n - M_{\text{nucleus}}$
Binding energy: $BE = \Delta m \times 931.5$ MeV
Radioactive decay: $N = N_0 e^{-\lambda t}$ , half-life $t_{1/2} = 0.693/\lambda$

The shortcut $E(\text{eV}) = 1240/\lambda(\text{nm})$ is the most time-saving formula in modern physics. It combines $h$ , $c$ , and the eV conversion into one number. Use it for every photoelectric and Bohr model numerical — it cuts calculation time by half.

Common Mistake

Students often write $KE_{\max} = h\nu + \phi$ instead of $h\nu - \phi$ . The work function is the energy spent to free the electron, so it is subtracted. If $KE_{\max}$ comes out negative, it means the photon does not have enough energy — no emission occurs. Also, the stopping potential in volts equals the maximum KE in eV numerically (since $eV_0 = KE_{\max}$ and $e$ cancels when KE is in eV).