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

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.