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Bohr model and photoelectric effect
Appears in
Concepts tested here
- Energy levels of hydrogen
- Photon energy and stopping potential
All Questions
2011 AIPMT-MAINS 1 question
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An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n isStopping potential 10 V means the maximum kinetic energy of the photoelectrons is 10 eV.
Energy of the incident photon: $E = KE_{max} + \phi_0 = 10 + 2.75 = 12.75$ eV
This photon is emitted in the transition from level n to the ground state (n = 1) of hydrogen: $E = 13.6\left(1 - \frac{1}{n^2}\right)$ eV
$12.75 = 13.6 - \frac{13.6}{n^2}$
$\frac{13.6}{n^2} = 13.6 - 12.75 = 0.85$
$n^2 = \frac{13.6}{0.85} = 16$
n = 4
2010 AIPMT-MAINS 1 question
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The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in $n^{th}$ state $E_n = -\frac{13.6}{n^2}$ eV)$E_3 = -\frac{13.6}{9} = -1.51$ eV and $E_1 = -13.6$ eV
Energy of the emitted photon: $h\nu = E_3 - E_1 = -1.51 - (-13.6) = 12.09$ eV $\approx 12.1$ eV
Einstein's equation: $KE_{max} = h\nu - \phi_0 = 12.1 - 5.1 = 7.0$ eV
$eV_0 = KE_{max}$
Stopping potential $V_0 = 7$ V