Question

Which of the following transition will emit maximum energy in hydrogen atom?
1. 4f----> 2s
2. 4d ----> 2p
3. 4p ---> 2s
4. all of these

Answer 1

4. All of these
Is the correct option
Because energy of the electron emitted depends only on the value of n, not on the orbital nature

Answer 2

Answer: Option (4) is the correct answer.

Explanation:

As it is known that relation between energy and number of shells is as follows.

                  Energy = $\frac{1}{n^{2}}$

where,   n = number of shell

As it is given that all the energy transition is taking place from n = 4 to n = 2.

So, in all the given cases energy will be as follows.

                 Energy = $\frac{1}{n^{2}}$

                              = $\frac{1}{2^{2}}$

                              = $\frac{1}{4}$

Therefore, we can conclude that all of these transitions will emit maximum energy in hydrogen atom.