Question

The work function of metal surface is 5 eV. If photon of wavelength 2480 Å strikes on the metal surface, then maximum kinetic energy (approx.) of photoelectron is

Answer 1

Step-by-step explanation:

Energy of incident radiation is E=

λ

hc

=

3×10

−7

6.6×10

−34

×3×10

8

=6.6×10

−19

J.

Also, work function is ϕ=2×1.6×10

−19

J=3.2×10

−19

J.

Hence, maximum kinetic energy of emitted photo electrons will be 6.6×10

−19

J−3.2×10

−19

J=3.4×10

−19

J.

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Answer 2

The kinetic energy of the photoelectron is zero.

Given,

Work function of the metal=5 eV

Wavelength of incident light=2480 Å.

To find,

the maximum kinetic energy (approx.) of photoelectron.

Solution:

• Work function(W) of a metal is the minimum energy that is required to surface electrons.
• It is measured in eV.
• When the energy of the incident light E is greater than the work function of the metal that it strikes, the electrons make best use of the incident energy and come out with kinetic energy.
• It is represented by the following expression:
• $KEmax=E-W$.
• This equation is known as Einstein's photoelectric equation.

It can further be written as:

$KEmax=\frac{hc}{lamda}-W$

where,

h-Plank's constant

$h=6.626X10^{-34} Jsec.$

c-speed of light

$c=3X10^{8} m/sec.$

λ-wavelength.

There is an another formula which can be used to find E or W

(in ev), it is given as:

$E(eV)=\frac{12400}{wavelength(in angstrom)} .$

The energy of the incident light will be:

$E(eV)=\frac{12400}{wavelength(in angstrom)} \\E=\frac{12400}{2480} \\E=5 eV.$

The kinetic energy of the photoelectron will be:

$KEmax=E-W\\KEmax=5-5\\KEmax=0.$

Hence, the kinetic energy is 0.

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