Single Choice

What will be the angular momentum of an electron, if the energy of this electron in H-atom is $$1.5 eV$$ (in J-s)?

A$$1.05\times { 10 }^{ -34 }$$
B$$2.1\times { 10 }^{ -34 }$$
C$$3.15\times { 10 }^{ -34 }$$
Correct Answer
D$$-2.1\times { 10 }^{ -34 }$$

Solution

Solution:- (C) $$3.15 \times {10}^{-34} {J}/{sec}$$
Energy of electron in $$H$$-atom $$\left( E \right) = \cfrac{-13.6}{{n}^{2}} \; eV$$
$$\Rightarrow \; {n}^{2} = \cfrac{-13.6}{E}$$
Given that energy of the electron in $$H$$-atom is $$1.5 \; eV$$.
$$\therefore \; {n}^{2} = \cfrac{-13.6}{- 1.5}$$
$$\Rightarrow \; n = \sqrt{9.067} \approx 3$$
As we know that, angular momentum $$\left( p \right) is given by-
$$p = \cfrac{nh}{2 \pi}$$
whereas, h is planck's constant $$= 6.6 \times {10}^{-34} {J}/{sec}$$
$$\therefore \; p = \cfrac{3 \times 6.6 \times {10}^{-34}}{2 \times 3.14} = 3.15 \times {10}^{-34} {J}/{sec}$$
Hence the angular momentum of electron will be $$3.15 \times {10}^{-34} \; {J}/{sec}$$.

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