Question

In which transition a hydrogen atom, photons of lowest frequency are emitted?
A. n=4 to n=3
B. n=4 to n=2
C. speed of an electron in the ${4}^{th}$ orbit of hydrogen
D. n=3 to n=1

Answer 1

Find the relation between energy and frequency. Use the formula for energy of photon due to transition of electron from one state of hydrogen atom to the other. Find the energy of each transition and then relate them with the relation you found between energy and frequency.

Complete answer:
Frequency of a photon emission is given by,
$\nu= \dfrac {\Delta E}{h}$ …(1)
Where, $\nu$ is the frequency of emission
h is the Planck’s constant
So, from the equation. (1) it can be inferred that frequency is directly proportional to energy of emission.
Thus, lower the energy of emission lower would be the frequency of the photon.
The formula for energy of photon due to transition from one energy state of hydrogen atom to the other is given by,
$E=R(\dfrac{1} {{n}_{1}^{2}}- \dfrac{1} {{n}_{2}^{2}})$
where, R is the Rydberg’s constant
${n}_{1}$ is the initial state
${n}_{2}$ is the final state

Substituting values in above equation from the options we get,
For n=4 to n=3,
$E=R(\dfrac{1} {{3}^{2}}- \dfrac{1} {{4}^{2}})$
$\Rightarrow E=R(\dfrac{1} {9}- \dfrac{1} {16})$
$\Rightarrow E=0.05R$

For n=4 to n=2
$E=R(\dfrac{1} {{2}^{2}}- \dfrac{1} {{4}^{2}})$
$\Rightarrow E=R(\dfrac{1} {4}- \dfrac{1} {16})$
$\Rightarrow E=0.08R$

For n=3 to n=1
$E=R(\dfrac{1} {{1}^{2}}- \dfrac{1} {{4}^{2}})$
$\Rightarrow E=R(\dfrac{1} {1}- \dfrac{1} {16})$
$\Rightarrow E=0.94R$
The lowest energy is for n=4 to n=3 transition.

So, the correct answer is “Option A”.

Note:
Remember that the ground state value for a hydrogen atom is 1 i.e. n=1.
The formula which defines energy levels of hydrogen atom is given by,
$E= -\dfrac {{E}_{0}}{{n}^{2}}$
where, ${E}_{0}$ is -13.6eV and n takes the value 1,2,… The energy is expressed as a negative number because it takes that much energy to unbind the electron from the nucleus.