(C) The energy of an electron in the $n^{\text{th}}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$. For the ground state $(n=1)$,$E_1 = -13.6 \ eV$

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(C) The energy of an electron in the $n^{\text{th}}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$. For the ground state $(n=1)$,$E_1 = -13.6 \ eV$. For the second orbit $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$. The energy of the emitted photon is the difference between these two states: $E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$. According to Einstein's photoelectric equation,$h\nu = \phi_0 + eV_s$,where $h\nu$ is the photon energy,$\phi_0$ is the work function,and $V_s$ is the stopping potential. Given $\phi_0 = 4.2 \ eV$ and $h\nu = 10.2 \ eV$,we have $10.2 \ eV = 4.2 \ eV + eV_s$. $eV_s = 10.2 \ eV - 4.2 \ eV = 6 \ eV$. Therefore,the stopping potential $V_s = 6 \ V$.

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

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An electron in a hydrogen atom jumps from the second Bohr orbit to the ground state,and the energy difference of the two states is radiated in the form of photons. These photons are then allowed to fall on a metal surface having a work function equal to $4.2 \ eV$. Calculate the stopping potential. [Energy of electron in $n^{\text{th}}$ orbit $= -\frac{13.6}{n^2} \ eV$] (in $V$)

MHT CET 2017 All MHT CET

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$2$
B
$4$
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$6$
D
$8$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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