Question 275: Dual Nature of Matter and Radiation

Question

Photoelectric emission takes place from a certain metal at threshold frequency $v$. If the radiation of frequency $4 v$ is incident on the metal plate, the maximum velocity of the emitted photoelectrons will be ( $m=$ mass of photoelectron, $h=$ Planck's constant)

Accepted Answer

Threshold frequency is given as $v$
Work function $\phi_0=\mathrm{h} 
u$
Now, $\mathrm{E}=$ K.E. $+\phi_0$
$$ \begin{array}{ll}  & 4 h v=\frac{1}{2} m v^2+h v \ 	herefore & \frac{1}{2} m v^2=4 h v-h v \ 	herefore & \frac{1}{2} m v^2=3 h v \ 	herefore & v=\sqrt{\frac{6 h v}{m}} \end{array} $$

Answer

$\sqrt{\frac{6 \mathrm{hv}}{\mathrm{m}}}$

Answer

$\sqrt{\frac{3 h v}{m}}$

Answer

$\sqrt{\frac{h v}{m}}$

Answer

$\sqrt{\frac{5 \mathrm{hv}}{\mathrm{m}}}$

Dual Nature of Matter and Radiation

Photoelectric emission takes place from a certain metal at threshold frequency v. If the radiation of frequency 4 v is incident on the metal plate, the maximum velocity of the emitted photoelectrons will be ( m= mass of photoelectron, h= Planck's constant)

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