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A photon of frequency 6.0 \times 10^{14} \mathrm{~Hz} is incident on a metal surface with work function 2.0 eV. What is the maximum kinetic energy of the emitted photoelectrons in eV?

Option: 1

0.4375 eV


Option: 2

2.0 eV


Option: 3

1.0 eV


Option: 4

3.5 eV


Answers (1)

best_answer

The energy of a photon is given by E=h \nu, where \mathrm{h} is Planck's constant and \nu is the frequency of the photon. In this case, the energy of the incident photon is
$$ \begin{gathered} E=h \nu=\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)(6.0 \times \\ \left.10^{14} \mathrm{~Hz}\right)=3.98 \times 10^{-19} \mathrm{~J} \end{gathered}
The work function of the metal is given as 2.0 \mathrm{eV}. We know that 1 \mathrm{eV} is equivalent to 1.6 \times 10^{-19} \mathrm{~J}. Therefore, the work function can be converted to joules as follows:
\begin{aligned} & \phi=2.0 \mathrm{eV} \times \frac{1.6 \times 10^{-19} \mathrm{~J}}{\mathrm{eV}}=3.2 \times \\ & 10^{-19} \mathrm{~J} \end{aligned}
The maximum kinetic energy of the photoelectrons can be found using the equation
K_{\max }=h \nu-\phi
Substituting the values, we get
K_{\max }=\left(6.626 \times 10^{-34} \mathrm{Js}\right) \cdot\left(6.0 \times 10^{14} \mathrm{~Hz}\right)-3.2 \times 10^{-19} \mathrm{~J}=0.7 \times 10^{-19} \mathrm{~J}

Converting this to electron volts (eV), we get

K_{\max }=\frac{0.7 \times 10^{-19}}{1.6 \times 10-19}=0.4375 \mathrm{eV}

Therefore, the maximum kinetic energy of the emitted photoelectrons is approximately 0.4375 eV.

 

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Divya Prakash Singh

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