Question

Photo electrons are emitted with a maximum speed of $7 \times 10^5 ms^{-1}$ 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. $4.64 Hz$

Answer 1

Answer: (B)

$h (\nu - \nu_0) = \frac{1}{2} mv^2_{max} $ or $\nu_0 = \nu - \frac{mv^2_{max}}{2h}$
= $8 \times 10^{14} - \frac{9.1 \times 10^{-31} \times 49 \times 10^{10}}{2 \times 6.63 \times 10^{-34}}$
= $8 \times 10^{14} - 3.36 \times 10^{14} = 4.64 \times 10^{14} $ Hz