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Question: An electron has mass \(9.1\, \times {10^{ - 28}}\) g and is moving with a velocity of \({10^5}\) cm/...

An electron has mass 9.1 ×10−289.1\, \times {10^{ - 28}} g and is moving with a velocity of 105{10^5} cm/sec. calculate its wavelength when h = 6.626× 10−276.626 \times \,{10^{ - 27}} erg-sec.
A.λ = 7.28 ×10−6 cm\lambda \, = \,7.28\, \times {10^{ - 6}}\,{\text{cm}}
B. λ = 7.28 ×10−5 cm\lambda \, = \,7.28\, \times {10^{ - 5}}\,{\text{cm}}
C.λ = 3.64 ×10−5 cm\lambda \, = \,3.64\, \times {10^{ - 5}}\,{\text{cm}}
D. λ = 3.64 ×10−6 cm\lambda \, = \,3.64\, \times {10^{ - 6}}\,{\text{cm}}

Explanation

Solution

We have to determine the wavelength from the given mass and velocity. So, we should know the relation among all. We can use the de-Broglie equation which represents the relation between wavelength, mass of the particle, velocity of the particle and Plank’s constant.

Complete solution:
The waves have a dual nature. In 19241924 De-Broglie told that the matter also has dual nature. The matter that has linear momentum also has a wave associated with it. The De-Broglie equation gives the relation between wave nature and particle nature. From the known wavelength its mass can be determined and from the given mass the wavelength can be determined.
De-Broglie equation is represented as follows:
λ= hmv\lambda = \,\dfrac{{\text{h}}}{{{\text{mv}}}}
λ\lambda is the de-Broglie wavelength of the particle.
h{\text{h}}is the Planck's constant.
m{\text{m}}is the mass of the particle.
v{\text{v}} is the velocity of the particle.
First we will convert the unit of h from erg to gram as follows:
11 erg = 1 gcm2s−2{\text{1}}\,{\text{gc}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 2}}
6.626× 10−276.626 \times \,{10^{ - 27}} erg. s = 6.626× 10−27 gcm2s−2.s6.626 \times \,{10^{ - 27}}\,{\text{gc}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 2}}.{\text{s}}
On substituting 9.1 ×10−289.1\, \times {10^{ - 28}} g for mass of electron, 105{10^5} cm/sec for velocity of electron, and 6.626× 10−27 gcm2s−2.s6.626 \times \,{10^{ - 27}}\,{\text{gc}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 2}}.{\text{s}}for h,
λ= hmv\lambda = \,\dfrac{{\text{h}}}{{{\text{mv}}}}
λ= 6.626× 10−27 gcm2s−2.s9.1 ×10−28 g × 105 cm s−1\lambda = \,\dfrac{{6.626 \times \,{{10}^{ - 27}}\,{\text{gc}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 2}}.{\text{s}}}}{{9.1\, \times {{10}^{ - 28}}\,{\text{g}}\, \times \,{{10}^5}\,{\text{cm}}\,{{\text{s}}^{ - 1}}}}
λ= 7.28 ×10−5 cm\lambda = \,7.28\, \times {10^{ - 5\,}}{\text{cm}}
So, an electron having mass 9.1 ×10−289.1\, \times {10^{ - 28}} g and is moving with a velocity of 105{10^5} cm/sec has wavelength equal to 7.28 ×10−5 7.28\, \times {10^{ - 5\,}} cm.

Therefore, option (B) is correct.

Note: The product of mass and velocity is given as momentum.so, the De-Broglie equation also written as λ= hp\lambda = \,\dfrac{{\text{h}}}{{\text{p}}}where, p{\text{p}}is the momentum. The de-Broglie wavelength of the particle is inversely proportional to its momentum. The unit of wavelength is meter. The unit of velocity is meter/second. h is known as Plank’s constant. The value of Planck's constant in joule is 6.6×10−34J s6.6 \times {10^{ - 34}}{\text{J}}\,{\text{s}}.