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A certain metal has a work function of 4.2 \mathrm{eV}. What is the maximum kinetic energy of the photoelectrons emitted when this metal is illuminated with ultraviolet light of wavelength 150 \mathrm{~nm} ? (Given: Planck's constant h=6.626 \times 10^{-34} \mathrm{Js}, speed of light =\left.c=3 \times 10^8 \mathrm{~m} / \mathrm{s}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)

Option: 1

0.58 \mathrm{eV}


Option: 2

0.21 \mathrm{eV}


Option: 3

2.48 \mathrm{eV}


Option: 4

4.20 \mathrm{eV}


Answers (1)

best_answer

$$ \lambda_0=\frac{h c}{\phi}=\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3 \times 10^8 \mathrm{~m} / \mathrm{s}}{4.2 \mathrm{eV}}= $$ 295 \mathrm{~nm}

Since the given wavelength (150 nm) is less than the threshold wavelength (295 nm), the photoelectric effect will occur and photoelectrons will be emitted.

The energy of a photon:

E=\frac{h c}{\lambda}

where E is the energy of the photon.

So, 

\begin{aligned} & E=\frac{6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3 \times 10^8 \mathrm{~m} / \mathrm{s}}{150 \times 10^{-9} \mathrm{~m}}= 4.41 \mathrm{eV} \end{aligned}

The kinetic energy \left(K_{\max }\right) of the photoelectrons is given by

 K_{\max }=E-\phi.
So,
K_{\max }=4.41 \mathrm{eV}-4.2 \mathrm{eV}=0.21 \mathrm{eV}
Therefore, the maximum kinetic energy of the photoelectrons emitted is 0.21 \mathrm{eV}.

Posted by

Rakesh

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