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  • If a metal has a work function of 4.5 eV, what is the minimum frequency of the incident light required to eject an electron from the metal surface?Option: 1</str

If a metal has a work function of 4.5 eV, what is the minimum frequency of the incident light required to eject an electron from the metal surface?

Option: 1

1.2\times10^{15}Hz


Option: 2

7.2\times10^{-20}Hz


Option: 3

2.25\times10^{-19}Hz


Option: 4

1.2\times10^{20}Hz


Answers (1)

best_answer

The energy required to eject an electron from a metal surface is given by the work function (\phi). This energy can also be expressed in terms of the frequency (f) of the incident light using the equation:

E= hf-\phi
where h is Planck's constant \left ( 6.626\times 10^{-34}J.s \right ).

To find the minimum frequency of the incident light required to eject an electron, we need to find the frequency that corresponds to the minimum energy required. This occurs when all of the energy of the incident light is used to eject the electron (i.e., hf= \phi).

Substituting the given work function of 4.5 eV into the equation gives:

E=\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)(f)-(4.5 \mathrm{eV})\left(1.602 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=0
Solving for f gives:
f=\frac{\phi}{h}=\frac{(4.5 \mathrm{eV})\left(1.602 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}} \approx 1.2 \times 10^{15} \mathrm{~Hz}


 

Posted by

shivangi.bhatnagar

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