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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)
A
5.1 V
B
12.1 V
C
17.2 V
D
7 V
Detailed Solution
$E_3 = -\frac{13.6}{9} = -1.51$ 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
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