Question

The angular momentum of an electron in a given stationary state can be expressed as ${m_e}vr = \dfrac{{nh}}{{2\pi }}$ . Based on this expression an electron can move only in those orbits for which its angular momentum is?
A.Equal to n
B.Integral multiple of $\dfrac{h}{{2\pi }}$
C.Multiple of n
D.Equal to $\dfrac{h}{{2\pi }}$ only.

Answer 1

We know that the number of orbits and their place around the nucleus is fixed. We also know that electrons can jump from one orbit to another orbit by emission or absorption of light energy. So their angular momentum must be quantized.

Complete answer:
The given expression gives a relation between angular momentum of a circulating particle and the ${n^{th}}$ orbital. This relation is derived from Bohr’s postulates which states that angular momentum of an electron in a given stationary state can be expressed as ${m_e}vr = \dfrac{{nh}}{{2\pi }}$ , where ${m_e}$ is mass of an electron, v is velocity of an electron, r is the radius of Bohr orbital and n is the ${n^{th}}$ Bohr orbit.
Since we know that only certain orbital energy levels are available around a nucleus of an atom so the quantity of those orbitals must be quantized. So an electron can revolve around a nucleus in only those orbits for which angular momentum of the electron is an integral multiple of $\dfrac{h}{{2\pi }}$ . Therefore n can be an integer such as 1,2,3 and so on.

So, option B is the correct answer.

Note:
Neils Bohr’s model was the first model to state that the orbits in which electrons revolve are quantized. His model was based on a hydrogen atom which contains only one electron but when larger atoms having more number of electrons were tested, his model failed.