Question

Light of wavelength 200 nm is incident on a photosensitive plate of work function 2 ev. Maximum kinetic energy of the photoelectron is

Answer 1

kmax=hc/lamda-W

Everything is given just equate .

Answer 2

Answer:

The Maximum kinetic energy of photoelectron is 2.959eV

Explanation:

Given that,

The wavelength of incident photon=200nm

Work function of photosensitive plate=2eV

According to Einstein's photoelectric equation,the energy of incident photon is sum of work function and maximum kinetic energy

$E=W+K.E$

Incident energy is found as follows

$E= \frac{hc}{λ} \\ = > E = \frac{6.626 \times 10 {}^{ - 34} \times 3 \times 10 {}^{8} }{200 \times 10 {}^{ - 9} } = \\ 9.939 \times 10 {}^{ - 19} joule$

Therefore,substituting this value in the Einstein equation,we get

$9.939 \times 10 {}^{ - 19} = 2 \times 1.609 \times 10 {}^{ - 19} + k.E \\ k.E = 9.939 \times 10 {}^{ - 19} - 2 \times 1.609 \times 10 {}^{ - 19} \\ = 6.721 \times 10 {}^{ - 19} joule \\ = 2.959eV$

The Maximum kinetic energy of photoelectron is 2.959eV

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