Question

Wavelength for high energy EMR transition in H-atom is $91nm$. What energy is needed for this transition?
$(A)1.36eV$
$(B)1240eV$
$(C)13eV$
$(D)13.6eV$

Answer 1

Hint: In order to solve this question, we will apply the formula of energy which is in terms of Planck’s constant, speed of light in vacuum and the wavelength. Then on putting the given value we will get the value of the wavelength and in the end, we will convert the value of wavelength from joule to electron volt.

Complete answer:
We know that the energy which is required for the transition of an electron in the H-atom is given by the expression,
$E = \dfrac{{hc}}{\lambda }$
Where,
$E$ is the energy required for the transition to take place
$h$ is the Planck’s constant
$c$ is the speed of light in vacuum
$\lambda $ is wavelength which is required for the transition to take place
In this question,
$h = 6.63 \times {10^{ - 34}}Js$
$c = 3 \times {10^8}\dfrac{m}{s}$
$\lambda = 91 \times {10^{ - 9}}m$
On putting the above values in the expression for the energy required for the transition to take place, we get,
$E = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{91 \times {{10}^{ - 9}}}}$
$E = \dfrac{{6.63 \times {{10}^{ - 17}} \times 3}}{{91}}$
On solving the above value, we get,
$E = 0.2186 \times {10^{ - 17}}$
$E = 2.186 \times {10^{ - 18}}J$
In the options given to us, we are given the value of energy in the form of electron volt. So, we need to convert the above value into electron volt. In order to convert it into electron volt, we will divide the above expression from the value of charge.
$E = \dfrac{{2.186 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}}$
$E = \dfrac{{2.186 \times 10}}{{1.6}}$
On further solving, we get,
$E = \dfrac{{21.86}}{{1.6}}$
$E = 13.6eV$
Thus, the energy required for the transition to take place is $E = 13.6eV$.

Therefore, the correct option is D

Note: The energy of an EMR transition is directly proportional to the frequency and is inversely proportional to the wavelength. So, we can say that the higher is the frequency of the EMR, the higher will be the energy and the higher is the wavelength of the EMR, the lower is the energy of the EMR.