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The radiation emitted, when an electron jumps from n=3 to n=2 orbit is a hydrogen atom, falls on a metal to produce photoelectron. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of \frac{1}{320}T in a radius of 10^{-3} m . Find the work function of metal-

Option: 1

1.03eV


Option: 2

1.89Ev


Option: 3

0.86 eV


Option: 4

2.03 eV


Answers (1)

best_answer

\begin{array}{r} E_3-E_2=13.6\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ \\\ =\frac{13.6 \times 5}{36}=1.89 \mathrm{eV} \end{array}
Photoelectrons with kE_{max} are moving on circular path.
\begin{aligned} & r=\frac{m v}{q B} \\ & m v=q B r \\ & P=q B r=1.6 \times 10^{-19} \times \frac{1}{3200} \times 10^{-3} \\ & \frac{1}{2} \times 10^{-24}=5 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \end{aligned}
Energy of photoelectron =KE _{max}= \frac{P^{2}}{2m}

\begin{aligned} & =\frac{25 \times 10^{-50}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}} \mathrm{eV} \\ \ & =0.86 \mathrm{eV} \end{aligned}
Now use Einstein equation
\begin{gathered} \mathrm{hv}=\phi+k E_{\max } \\ 0.56+\phi ; \quad \phi=1.03 \mathrm{ev} \end{gathered}

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