Q85 / 120 · VITEEE 2008 · Physics
A light whose frequency is equal to $6 \times 10^{14} \mathrm{~Hz}$ is incident on a metal whose work function is
$2 \mathrm{eV} \cdot\left[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\right]\left[1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right]$
The maximum energy of the electrons emitted will be
- A$2.49 \mathrm{eV}$
- B$4.49 \mathrm{eV}$
- Correct answerC$0.49 \mathrm{eV}$
- D$5.49 \mathrm{eV}$
Solution
Here,
Frequency, $v=6 \times 10^{14} \mathrm{~Hz}$
Work-function, $\phi=2 \mathrm{eV}=2 \times 1.6 \times 10^{-19} \mathrm{~J}$
$=3.2 \times 10^{-19} \mathrm{~J}$
Maximum energy, $\mathrm{T}_{\max }=$ ?
By Einstein's photo electric equation, we have
$\begin{aligned} & h v=\phi+T_{\max } \\ \Rightarrow & T_{\max }=h v-\phi \\=&\left(6.63 \times 10^{-34} \times 6 \times 10^{14}\right)-\left(3.2 \times 10^{-19}\right) \\=&\left(3.97 \times 10^{-19}\right)-\left(3.2 \times 10^{-19}\right) \\=& 0.77 \times 10^{-19} \mathrm{~J} \\=& \frac{0.77 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=0.49 \mathrm{eV} \end{aligned}$
Hence, the correct answer is option: 'C'.