Question

An electron of energy $ 150eV $ has a wavelength of $ {10^{ - 10}}m $ . The wavelength of a $ 0.60keV $ electron is:
 $ \left( A \right){\text{ 0}}{\text{.50}}{\kern 1pt} \mathop A\limits^ \circ $
 $ \left( B \right){\text{ 0}}{\text{.75}}{\kern 1pt} \mathop A\limits^ \circ $
 $ \left( C \right){\text{ 1}}{\text{.2}}{\kern 1pt} \mathop A\limits^ \circ $
 $ \left( D \right){\text{ 1}}{\text{.5}}\mathop A\limits^ \circ $

Answer 1

Hint: Here in this question we have to find the wavelength of the other electron so for this we will use the equation of de Broglie, and relation is given by $ \dfrac{h}{{\sqrt {meV} }} $ . So by using this equation we will get the relation and then can solve for the value of other lambda we will get the answer.

Formula used:
On the basis of de Broglie relation,
 $ \lambda = \dfrac{h}{{\sqrt {meV} }} $
Here, $ \lambda $ will be the wavelength
 $ h $ , will be the Planck’s constant
 $ m $ , will be the mass
 $ V $ , will be the velocity.

Complete step by step solution:
Since, we know that the mass and the Planck;s constant is constant in the equation of de Broglie. So from this wavelength will vary as $ \dfrac{1}{{\sqrt V }} $ .
So for the first wavelength the equation will be,
 $ \Rightarrow {\lambda _1}\propto \dfrac{1}{{\sqrt {{V_1}} }} $
And for the second wavelength it is given by,
 $ \Rightarrow {\lambda _2}\propto \dfrac{1}{{\sqrt {{V_2}} }} $
Therefore from the above two relation, the ratio of lambda will be equal to
 $ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\dfrac{{{V_2}}}{{{V_1}}}} $
Now by substituting the values, we will get the equation as
 $ \Rightarrow \dfrac{{{{10}^{ - 10}}}}{{{\lambda _2}}} = \sqrt {\dfrac{{0.60 \times {{10}^3}}}{{150}}} $
And on solving the RHS of the equation we will get the equation as
 $ \Rightarrow \dfrac{{{{10}^{ - 10}}}}{{{\lambda _2}}} = \sqrt {\dfrac{{600}}{{150}}} $
Now on solving the equation, we get
 $ \Rightarrow {\lambda _2} = 0.5 \times {10^{ - 10}}m $
And it can also be written as
 $ \Rightarrow {\lambda _2} = 0.5\mathop A\limits^ \circ $
Hence, the wavelength of a $ 0.60keV $ electron is $ 0.5\mathop A\limits^ \circ $
Therefore, the option $ \left( A \right) $ is correct.

Note:
De Broglie wavelength is a wavelength, which is manifested in all the particles in quantum mechanics, according to wave-particle duality, and it determines the probability density of finding the object at a given point of the shape space.