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Question

(B) The energy of the photon emitted when an electron jumps from the first excited state $(n=2)$ to the ground state $(n=1)$ in a hydrogen atom is giv

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

(B) The energy of the photon emitted when an electron jumps from the first excited state $(n=2)$ to the ground state $(n=1)$ in a hydrogen atom is given by:$E = E_2 - E_1 = -3.4 \; eV - (-13.6 \; eV) = 10.2 \; eV$.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.Given $V_s = 3.57 \; V$,so $K_{max} = 3.57 \; eV$.Substituting the values:$10.2 \; eV = \phi + 3.57 \; eV$$\phi = 10.2 - 3.57 = 6.63 \; eV$.The work function is also given by $\phi = h 
u_0$,where $h \approx 4.136 	imes 10^{-15} \; eV \cdot s$.$
u_0 = \frac{\phi}{h} = \frac{6.63 \; eV}{4.136 	imes 10^{-15} \; eV \cdot s} \approx 1.6 	imes 10^{15} \; Hz$.

Monochromatic radiation emitted when an electron in a hydrogen atom jumps from the first excited state to the ground state irradiates a photosensitive material. The stopping potential is measured to be $3.57 \; V$. The threshold frequency of the material is ......... $\times 10^{15} \; Hz$.

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