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The threshold frequency for a certain photosensitive metal is $\mathrm{\nu}_0$ . When it is illuminated by light of frequency $\mathrm{\nu}=2 \mathrm{\nu}_0$ , the maximum velocity of photoelectrons is $\mathrm{\nu}_0$ . What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency $\nu=5 \nu_0$ ?
Option: 1
$\sqrt{2} \mathrm{\nu}_0$

Option: 2
$2 \mathrm{\nu}_0$

Option: 3
$2 \sqrt{2} \mathrm{\nu}_0$

Option: 4
$4 \nu_0$

Answers (1)

best_answer

As $\mathrm{\nu}_0$ is the threshold frequency.

$\therefore$ Work function, $\phi_0=h \nu_0$

According to Einstein photoelectric equation

$\frac{1}{2} \mathrm{mv}_{\max }^2=\mathrm{h\nu}-\phi_0$

Where $\mathrm{h\nu}$ is the incident energy, $\phi_0$ is the work function of the metal and $\frac{1}{2} \mathrm{mv}_{\max }^2$ is the maximum kinetic energy of the emitted photoelectron.

As per question
$\frac{1}{2} m v_0^2=h\left(2 \nu_0\right)-h \nu_0=h \nu_0 \quad \quad \quad \quad (i)$
and $\frac{1}{2} \mathrm{mv}^{\prime 2}=5\left(5 \mathrm{\nu}_0\right)-h \mathrm{\nu}_0=4 h \mathrm{\nu}_0 \quad \quad \quad \quad (ii)$

Divide (ii) by (i), and we get

$\begin{array}{ll} \frac{\nu^{\prime 2}}{\nu_0^2}=\frac{4}{1} & \\ \\ \nu^{\prime 2}=4 \nu_0^2 & \text { or } \quad \nu^{\prime}=2 \nu_0 \end{array}$

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