Physics, asked by ishapurandare, 1 year ago

a metal whose work function is 4.2 eV is irradiated by radiation whose wavelength is 2000 Armstrong. find the maximum kinetic energy of emitted electron

Answers

Answered by Geekydude121
17
According to question

Given that
work function = 4.2 eV
We know 
Planks Constant
hf = 6.63 X 10^-34
And Also
Minimum Kinetic Energy

Ek = hf - work function
      = 6.63 X 10^-34 - 4.2 eV
       = 2.43 J
Thus the minimum Kinetic Energy is 2.43 J
Answered by muscardinus
5

The maximum kinetic energy of emitted electron is 3.225\times 10^{-19}\ J

Explanation:

It is given that,

The work energy of a metal, \phi=4.2\ eV=4.2\times 1.6\times 10^{-19}\ J=6.72\times 10^{-19}\ J

Wavelength of the radiation, \lambda=2000\ A=2\times 10^{-7}\ m

According to Einstein's photoelectric equation,

hf=E_{max}+\phi

h\dfrac{c}{\lambda}=E_{max}+\phi

E_{max}=h\dfrac{c}{\lambda}-\phi

E_{max}=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{2\times 10^{-7}}-6.72\times 10^{-19}

E_{max}=3.225\times 10^{-19}\ J

So, the maximum kinetic energy of emitted electron is 3.225\times 10^{-19}\ J. Hence, this is the required solution.

Learn more :

Einstein's equation of photoelectric effect

https://brainly.in/question/10797281

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