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Threshold wavelength for photoelectric emission from a metal surface is $5200 Å$. Photoelectrons will be emitted when this surface is illuminated with monochromatic radiation from
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The correct answer is:
$50 \mathrm{~W} \mathrm{UV}-\operatorname{lamp}$
Threshold wavelength of the metal, $\lambda_{0}=5200 Å$
$$
=5200 \times 10^{-10} \mathrm{~m}
$$
Frequency of the radiation, $v_{0}=\frac{c}{\lambda}$
$$
\begin{aligned}
&=\frac{3 \times 10^{8}}{5200 \times 10^{-10}} \\
&=057 \times 10^{15} \mathrm{~Hz} \\
&=5.7 \times 10^{14} \mathrm{~Hz}
\end{aligned}
$$
The frequency of infrared radiation is lower than the threshold frequency of the metal. Hence, the surface of the metal is illuminated with monochromatic radiation which is greater than $5.7 \times 10^{14} \mathrm{~Hz}$.
The $50 \mathrm{~W}$ LV-lamp is suitable for this purpose.
$$
=5200 \times 10^{-10} \mathrm{~m}
$$
Frequency of the radiation, $v_{0}=\frac{c}{\lambda}$
$$
\begin{aligned}
&=\frac{3 \times 10^{8}}{5200 \times 10^{-10}} \\
&=057 \times 10^{15} \mathrm{~Hz} \\
&=5.7 \times 10^{14} \mathrm{~Hz}
\end{aligned}
$$
The frequency of infrared radiation is lower than the threshold frequency of the metal. Hence, the surface of the metal is illuminated with monochromatic radiation which is greater than $5.7 \times 10^{14} \mathrm{~Hz}$.
The $50 \mathrm{~W}$ LV-lamp is suitable for this purpose.
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