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When a certain metal was irradiated with light of frequency $3.2 \times 10^{16} H z$ , the photoelectrons emitted had twice the kinetic energy as did photoelectrons when the same metal was irradiated with light of frequency $2.0 \times 10^{16} H z$ . The threshold frequency for the metal is

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$1.2 \times 10^{16} H z$

$8 \times 10^{15} H z$

$8 \times 10^{6} H z$

$1.2 \times 10^{15} H z$

answer is B.

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Detailed Solution

$h ϑ_{1} = h ϑ_{0} + K E_{1}$

$h ϑ_{2} = h ϑ_{0} + K E_{2}$

$\frac{h (ϑ_{1} - ϑ_{0})}{h (ϑ_{2} - ϑ_{0})} = \frac{E_{1}}{E_{2}}$

$\frac{3.2 \times 10^{16} - ϑ_{0}}{2.0 \times 10^{16} - ϑ_{0}} = \frac{2}{1}$

$3.2 \times 10^{16} - ϑ_{0} = 2 (2.0 \times 10^{16} - ϑ_{0})$

$3.2 \times 10^{16} - ϑ_{0} = 4 \times 10^{16} - 2 ϑ_{0}$

$ϑ_{0} = 0.8 \times 10^{16}$

$\begin{matrix} ϑ_{0} = 8 \times 10^{15} \end{matrix}$

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