Threshold frequency for a metal is $15 \times 10^{14} \mathrm{~Hz}$. The light of wavelength $6000 Å$ falls…
Threshold frequency for a metal is $15 \times 10^{14} \mathrm{~Hz}$. The light of wavelength $6000 Å$
falls on the metal surface. Which one of the following statements is correct?
[velocity of light, $\left.\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right]$
photoelectrons are emitted with velocity c.
photoelectrons come out with velocity $3 \times 10^{6} \mathrm{~m} / \mathrm{s}$
photoelectrons come out with zero velocity.
photoelectrons will not be emitted.
Solution
$\begin{aligned}
v_{0} &=15 \times 10^{14} \mathrm{~Hz} \\
\therefore \lambda_{0}=\frac{C}{v} &=\frac{3 \times 10^{8}}{15 \times 10^{14}} \\
&=2 \times 10^{-7} \mathrm{~m} \\
&=2000 Å \\
\lambda &=6000 Å
\end{aligned}$
Since $\lambda>\lambda_{0}$, photo electrons will not be emitted