Question

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x10^1-7 J (A) 2.18x10- (B) 7.84x10- (C) 8.62x10-13 (D) 5.34x10-5

Answer 1

Answer : The kinetic energy acquired by the electron in hydrogen atom is $7.12\times 10^{-22}J$

Explanation :

First we have to calculate the wavelength.

Formula used :

$E=\frac{hc}{\lambda}$

where,

E = energy = $1.08\times 10^{-17}J$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = ?

Now put all the given values in the above formula, we get:

$1.08\times 10^{-17}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}$

$\lambda=1.84\times 10^{-8}m$

Now we have to determine the kinetic energy acquired by the electron in hydrogen atom.

According to de-Broglie, the expression for wavelength is,

$\lambda=\frac{h}{p}$

and,

$p=mv$

where,  

p = momentum, m = mass, v = velocity

So, the formula will be:

$\lambda=\frac{h}{mv}$

or,

$v=\frac{h}{m\lambda}$     ............(1)

The formula of kinetic energy is:

$K.E=\frac{1}{2}mv^2$      ..........(2)

Now put the equation 1 in 2, we get:

$K.E=\frac{1}{2}m(\frac{h}{m\lambda})^2$

$K.E=\frac{1}{2}(\frac{h^2}{m\lambda^2})$         ............(3)

where,

m = mass of electron = $9.1\times 10^{-31}kg$

Now put all the given values in the formula 3, we get:

$K.E=\frac{1}{2}(\frac{(6.626\times 10^{-34}Js)^2}{(9.1\times 10^{-31}kg)\times (1.84\times 10^{-8}m)^2})$

$K.E=7.12\times 10^{-22}J$

Therefore, the kinetic energy acquired by the electron in hydrogen atom is $7.12\times 10^{-22}J$