Question

Photo electrons are emitted with a maximum speed of $7\times 10^{5}m{s}^{-1}$ from a surface when light of frequency $8\times 10^{14}Hz$ is incident on it, the threshold frequency for this surface is

• A. $2.32\times 10^{14}Hz$
• B. $4.64\times 10^{14}Hz$
• C. $4.64\times 10^{20}Hz$
• D. $4.64Hz$

Answer 1

Answer: (B)

$h(\nu -{\nu }_0)=\frac{1}{2}m{v}_{max}^2$ or ${\nu }_0=\nu -\frac{m{v}_{max}^2}{2h}$
= $8\times 10^{14}-\frac{9.1\times 10^{-31}\times 49\times 10^{10}}{2\times 6.63\times 10^{-34}}$
= $8\times 10^{14}-3.36\times 10^{14}=4.64\times 10^{14}$ Hz