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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 is
A
5
B
2
C
3
D
4
Detailed Solution
Stopping 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
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