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The photosensitive metallic surface has work function $h v_0$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^6 \mathrm{~m} / \mathrm{s}$. When the photon energy is increased $5 h v_0$. Then maximum velocity of photoelectrons will be:
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Verified Answer
The correct answer is:
$8 \times 10^6 \mathrm{~m} / \mathrm{s}$
According to the question
$$
\begin{gathered}
2 h v_0=h v_0+\frac{1}{2} m\left(4 \times 10^6\right)^2 \\
5 h v_0=h v_0+\frac{1}{2} m v^2 \\
\Rightarrow 4 \times \frac{1}{2} m\left(4 \times 10^6\right) \\
\Rightarrow \frac{1}{2} m v^2 \\
v=8 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$
$$
\begin{gathered}
2 h v_0=h v_0+\frac{1}{2} m\left(4 \times 10^6\right)^2 \\
5 h v_0=h v_0+\frac{1}{2} m v^2 \\
\Rightarrow 4 \times \frac{1}{2} m\left(4 \times 10^6\right) \\
\Rightarrow \frac{1}{2} m v^2 \\
v=8 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$
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