Question.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency $\left(\begin{array}{ll}v_{0} & )\end{array}\right.$ and work function (W0) of the metal.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency $\left(\begin{array}{ll}v_{0} & )\end{array}\right.$ and work function (W0) of the metal.
Solution:
Threshold wavelength of radiation $\left(\lambda_{0}\right)=6800 =6800 \times 10^{-10} \mathrm{~m}$
Threshold frequency $\left(v_{0}\right)$ of the metal
$=\frac{c}{\lambda_{0}}=\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{6.8 \times 10^{-7} \mathrm{~m}}=4.41 \times 10^{14} \mathrm{~s}^{-1}$
Thus, the threshold frequency $\left(v_{0}\right)$ of the metal is $4.41 \times 10^{14} \mathrm{~s}^{-1}$.
Hence, work function $\left(W_{0}\right)$ of the metal $=h v_{0}$
$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(4.41 \times 10^{14} \mathrm{~s}^{-1}\right)$
$\left.=2.922 \times 10^{-19}\right]$
Threshold wavelength of radiation $\left(\lambda_{0}\right)=6800 =6800 \times 10^{-10} \mathrm{~m}$
Threshold frequency $\left(v_{0}\right)$ of the metal
$=\frac{c}{\lambda_{0}}=\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{6.8 \times 10^{-7} \mathrm{~m}}=4.41 \times 10^{14} \mathrm{~s}^{-1}$
Thus, the threshold frequency $\left(v_{0}\right)$ of the metal is $4.41 \times 10^{14} \mathrm{~s}^{-1}$.
Hence, work function $\left(W_{0}\right)$ of the metal $=h v_{0}$
$=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(4.41 \times 10^{14} \mathrm{~s}^{-1}\right)$
$\left.=2.922 \times 10^{-19}\right]$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.