Question

The energy that should be added to an electron to reduce its De Broglie wavelength from $1\,nm$ to $0.5\,nm$ is.
A. Four times the initial energy
B. Equal to initial energy
C. Twice the initial energy
D. Thrice the initial energy

Answer 1

In order to solve this question we need to understand the debuggable wavelength. According to de broglie, a wave is associated with every matter known as matter waves, so wavelength of matter waves depends inversely on the momentum of an object or matter. It is mathematically defined as, ratio of Planck constant and momentum of matter. In this question we would first relate debroglie wavelength to energy of matter and then compare for two cases.

Complete step by step answer:
Let “$M$” be the mass of particle and also let us assume that it is moving with velocity “$v$”
Then momentum of particle is defined as, $p = Mv \to (i)$
Also, kinetic energy of the particle is $E = \dfrac{1}{2}M{v^2}$
From equation (i) value of velocity is, $v = \dfrac{p}{M}$
So putting the value of velocity in kinetic energy expression we get,
$E = \dfrac{1}{2}M{(\dfrac{p}{M})^2}$
$\Rightarrow E = \dfrac{{{p^2}}}{{2M}}$
So the momentum is given in terms of energy as,
${p^2} = 2ME$
$\Rightarrow p = \sqrt {2ME} \to (ii)$

Let the De Broglie wavelength be $\lambda$.So according to de broglie, wavelength is defined as,
$\lambda = \dfrac{h}{p}$.
Putting value of momentum from equation (ii) we get,
$\lambda = \dfrac{h}{{\sqrt {2ME} }}$
So energy in terms of wavelength is given as,
$E = \dfrac{{{h^2}}}{{2M{\lambda ^2}}} \to (iii)$
So for case $1$ we have ${\lambda _1} = 1nm$
So first energy from equation (iii) is given as,
${E_1} = \dfrac{{{h^2}}}{{2M{\lambda _1}^2}}$
And for case $2$ we have ${\lambda _2} = 0.5\,nm$

So first energy from equation (iii) is given as,
${E_2} = \dfrac{{{h^2}}}{{2M{\lambda _2}^2}}$
Dividing both energies we get,
$\dfrac{{{E_2}}}{{{E_1}}} = \dfrac{{\dfrac{{{h^2}}}{{2M{\lambda _2}^2}}}}{{\dfrac{{{h^2}}}{{2M{\lambda _1}^2}}}}$
$\Rightarrow \dfrac{{{E_2}}}{{{E_1}}} = \dfrac{{{\lambda _1}^2}}{{{\lambda _2}^2}}$
Putting values we get,
$\dfrac{{{E_2}}}{{{E_1}}} = {(\dfrac{1}{{0.5}})^2}$
$\Rightarrow \dfrac{{{E_2}}}{{{E_1}}} = 4$
$\therefore {E_2} = 4{E_1}$

Therefore, the correct option is A.

Note: It should be remembered that, here we have related both momentum and energy by assuming non relativistic case that is speed of electron is very less in comparison to light speed, however if we consider relativistic case then it would not be direct relation rather rest mass energy also come into consideration.