Question

What is E of a hydrogen atom in the 3p state?

Answer 1

Answer:

E=-1.51 eV.

$L=\hbar\sqrt{2}$

Explanation:

The nth level energy of a hydrogen atom is defined by the formula,

$E_{n}=-\frac{13.6}{n^{2} }$

Given in the question, the hydrogen atom is in the 3p state.

Then energy of n=3 state is,

$E_{n}=-\frac{13.6}{(3)^{2} }\\E_{n}=-1.51eV$

Therefore, energy of the hydrogen atom in the 3p state is -1.51 eV.

Now, the value of L can be calculated as,

$L=\hbar\sqrt{l(l+1)}$

For 3p state, l=1

$L=\hbar\sqrt{1(1+1)}\\L=\hbar\sqrt{2}$

Therefore, the value of L of a hydrogen atom in 3p state is $L=\hbar\sqrt{2}$.