Question

3. Light of frequency 5 x 1014 HZ liberates electrons with energy 2.3 x 10-19from a certain metallic surface. What is the wavelength of ultraviolet light which liberates electrons of energy 8.93 x 10-19 J from the same surface?​

Answer 1

Answer: lambda is 200nm

Explanation:hf =ke.+work function
6.63×10^-34×5×10^14=2.31×10^-19+work function
Work function=1.005×10^-19
hf =8.93×10-19 + 1.005×10^-19
hf=9.935×10-19
hc÷lamda=9.935×10-19
Lambda =6.63×10-34×3×10^8÷9.935×10-19
Lambda=2.0×10^7m
=200nm

Answer 2

The wavelength of the ultraviolet light is 0.6 micro-metres.

Given,

Frequency, f=$5X10^{14}$ Hertz

Energy of the electrons, E1=$2.3X10^{-19}$J

Energy of the electrons, E2=$8.93X10^{-19}$J.

To find,

the wavelength of the ultraviolet light.

Solution:

• As the incident light strikes the same metallic surface then the work function will be same.
• 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.

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

This equation is known as Einstein's photoelectric equation.

It can be further written as:

$W=KEmax-\frac{hc}{ λ}$

where,

h-Plank's constant

$h=6.626X10^{2-34} J sec$

c-speed of light

$c=3X10^{8} m/s$

λ-wavelength

Assuming that the K.E. energy is same in both the cases(as there is no information about the velocity),

$W1=W2\\KEmax-E1=KEmax-E2\\E1=E2\\hf=\frac{hc}{ λ} \\f=\frac{c}{ λ} \\5X10^{14} =\frac{3X10^{8} }{ λ} \\ λ=\frac{3X10^{8} }{5X10^{14} } \\ λ=0.6X10^{-6} m\\ λ=0.6 micro- m.$

($1 micro-m=10^{-6} m$)

The wavelength is 0.6 micro-metres.

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