Question

When a certain metal was irradiated with light of frequency $1.6\times {10}^{16}Hz$, the photoelectrons emitted had twice the kinetic energy as the photoelectrons emitted when the same metal was irradiated with light of frequency $1.0\times {10}^{16}Hz$. Calculate the threshold frequency ($v_0$) for the metal.

• A. 2 x 10¹⁵ Hz
• B. 6 x 10¹⁵ Hz
• C. 12 x 10¹⁵ Hz
• D. 4 x 10¹⁵ Hz

Answer 1

The correct option is D

4 x 10¹⁵ Hz

From Einstein's equation, we know

KE = $h\nu$ - $h{\nu }_0$

$K{E}_2$ = $\frac{K{E}_1}{2}$

Now, $K{E}_1$ = $h(\nu -{\nu }_0)$

$K{E}_2$ = $h({\nu }_2-{\nu }_0)$

$\Rightarrow \frac{{\nu }_2-{\nu }_0}{{\nu }_1-{\nu }_0}$ = $\frac{1}{2}$

$\Rightarrow$$\frac{1\times {10}^{16}-{\nu }_0}{1.6\times {10}^{16}-{\nu }_0}$ = $\frac{1}{2}$

$\Rightarrow Thethresholdfrequency$ ${\nu }_0$ = $4\times {10}^{15}Hz$