Question

A certain metal was irradiated with light of frequency $3.2\times {10}^{16}$ Hz. The photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency $2\times {10}^{16}$Hz. Calculate the threshold frequency for the metal.

• A. $8\times {10}^{15}Hz$
• B. $6\times {10}^{15}Hz$
• C. $8\times {10}^{12}Hz$
• D. $6\times {10}^{12}Hz$

Answer 1

The correct option is A $8\times {10}^{15}Hz$

Given frequency, ${\nu }_2=3.2\times {10}^{16}$ Hz.
${\nu }_1=2\times {10}^{16}$ Hz.
We know that from photoelectric effect equation
$KE.=h\nu -h{\nu }_{0}or\nu -{\nu }_0=\frac{K.E}{h}$
$K{E}_2=2K{E}_1$
${\nu }_2-{\nu }_0=\frac{K{E}_2}{h};{\nu }_1-{\nu }_0=\frac{K{E}_1}{h}$
Dividing these equations yields $\frac{{\nu }_2-{\nu }_0}{{\nu }_1-{\nu }_0}=\frac{K{E}_{2/h}}{K{E}_1/h}=2$
$\therefore {\nu }_2-{\nu }_0=2{\nu }_1-2{\nu }_0$
$\therefore {\nu }_0=2{\nu }_1-{\nu }_2=2(2\times {10}^{16})-(3.2\times {10}^{16})=8\times {10}^{15}Hz$
Hence, the threshold frequency for the metal is $8\times {10}^{15}Hz$