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A light of frequency $1.6 \times 10^{16} \mathrm{~Hz}$ when falls on a metal plate emits electrons that have double the kinetic energy compared to the kinetic energy of emitted electrons when frequency of $1.0 \times 10^{16} \mathrm{~Hz}$ falls on the same plate. The threshold frequency $\left(v_0\right)$ of the metal in $\mathrm{Hz}$ is
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Verified Answer
The correct answer is:
$4 \times 10^{15}$
When $1.6 \times 10^{16} \mathrm{~Hz}$ frequency falls on metal plate than double kinetic energy obtained.
$$
h\left(1.6 \times 10^{16}-v_0\right)=2 \mathrm{~K} . \mathrm{E} \longrightarrow \text { Eq. (i) }
$$
When $1.0 \times 10^{16} \mathrm{~Hz}$ frequency falls on same plate than K.E.
$$
h\left(1.0 \times 10^{16}-v_0\right)=\text { K.E } \longrightarrow(\text { Eq. (ii) })
$$
From Eqs. (i) and (ii)
$$
\begin{aligned}
& \frac{h\left(1.6 \times 10^{16}-v_0\right)}{h\left(1.0 \times 10^{16}-v_0\right)}=\frac{2 \mathrm{~K} . \mathrm{E}}{\mathrm{K} . \mathrm{E}} \\
& 1.6 \times 10^{16}-v_0=2 \times 1.0 \times 10^{16}-v_0 \\
& 1.6 \times 10^{16}-2\left(1.0 \times 10^{16}\right)=4 \times 10^{15} \mathrm{~Hz} \\
& v_0=4 \times 10^{15} \mathrm{~Hz}
\end{aligned}
$$
$$
h\left(1.6 \times 10^{16}-v_0\right)=2 \mathrm{~K} . \mathrm{E} \longrightarrow \text { Eq. (i) }
$$
When $1.0 \times 10^{16} \mathrm{~Hz}$ frequency falls on same plate than K.E.
$$
h\left(1.0 \times 10^{16}-v_0\right)=\text { K.E } \longrightarrow(\text { Eq. (ii) })
$$
From Eqs. (i) and (ii)
$$
\begin{aligned}
& \frac{h\left(1.6 \times 10^{16}-v_0\right)}{h\left(1.0 \times 10^{16}-v_0\right)}=\frac{2 \mathrm{~K} . \mathrm{E}}{\mathrm{K} . \mathrm{E}} \\
& 1.6 \times 10^{16}-v_0=2 \times 1.0 \times 10^{16}-v_0 \\
& 1.6 \times 10^{16}-2\left(1.0 \times 10^{16}\right)=4 \times 10^{15} \mathrm{~Hz} \\
& v_0=4 \times 10^{15} \mathrm{~Hz}
\end{aligned}
$$
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