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If the maximum kinetic energy of emitted electrons in photoelectric effect is $3 \cdot 2 \times 10^{-19} \mathrm{~J}$ and the work function for metal is $6 \cdot 63 \times 10^{-19} \mathrm{~J}$, then stopping
potential and threshold wavelength respectively are
[Planck's constant $\left.\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right]$
[Velocity of light $\left.\mathrm{c}=3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}}\right]$
$\left[\right.$ charge on electron $\left.=1 \cdot 6 \times 10^{-19} \mathrm{C}\right]$
Options:
potential and threshold wavelength respectively are
[Planck's constant $\left.\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right]$
[Velocity of light $\left.\mathrm{c}=3 \times 10^{8} \frac{\mathrm{m}}{\mathrm{s}}\right]$
$\left[\right.$ charge on electron $\left.=1 \cdot 6 \times 10^{-19} \mathrm{C}\right]$
Solution:
2131 Upvotes
Verified Answer
The correct answer is:
$2 \mathrm{~V}, 3000 Å$
(D)
$\begin{array}{l}
(\mathrm{k} \cdot \mathrm{E})_{\max }=\mathrm{eV}_{\mathrm{s}}=3.2 \times 10^{-19} \mathrm{~J} \\
\therefore \mathrm{V}_{\mathrm{s}}=\frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}}=2 \mathrm{~V}
\end{array}$
In the given options, only option (D) has $\mathrm{V}_{\mathrm{s}}=2 \mathrm{~V}$
$\begin{array}{l}
(\mathrm{k} \cdot \mathrm{E})_{\max }=\mathrm{eV}_{\mathrm{s}}=3.2 \times 10^{-19} \mathrm{~J} \\
\therefore \mathrm{V}_{\mathrm{s}}=\frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}}=2 \mathrm{~V}
\end{array}$
In the given options, only option (D) has $\mathrm{V}_{\mathrm{s}}=2 \mathrm{~V}$
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