(C) The energy of the photon emitted during the transition from $n=2$ to $n=1$ is given by:$E = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) =

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(C) The energy of the photon emitted during the transition from $n=2$ to $n=1$ is given by:$E = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) = -3.4 + 13.6 = 10.2 \ eV$.According to Einstein's photoelectric equation:$E = \phi + K_{max}$, where $\phi$ is the work function and $K_{max} = eV_s$ is the maximum kinetic energy.$10.2 \ eV = 4.2 \ eV + eV_s$.$eV_s = 10.2 \ eV - 4.2 \ eV = 6.0 \ eV$.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 difference between the energies of the two states is radiated in the form of a photon. This photon strikes a material. If the work function of the material is $4.2 \ eV$, then the stopping potential is (Energy of electron in $n$-th orbit $= -\frac{13.6}{n^2} \ eV$). (in $V$)

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A
$2$
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$4$
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$6$
D
$8$

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Class 12

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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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In the graph given below,if the slope is $4.12 \times 10^{-15} \ V-sec$,then the value of $h$ should be:

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Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

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