An electron in an excited state of $\mathrm{Li}^{2+}$ ion has angular momentum $3 h / 2 \pi$. The de Broglie wavelength of the electron in this state is $p \pi a_{0}$ (where $a_{0}$ is the Bohr radius). The value of $p$ is

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Modern Physics

An electron in an excited state of

Li^{2 +}

ion has angular momentum

3 h /2 π

. The de Broglie wavelength of the electron in this state is

p π a_{0}

(where

a_{0}

is the Bohr radius). The value of

p

Answer: 2

Solution

Given angular momentum

L = \frac{nh}{2 π} = \frac{3 h}{2 π}

, so

n = 3

.

The de Broglie wavelength is:

λ = \frac{h}{m v} = \frac{h \cdot 2 π r}{nh} = \frac{2 π r}{n}

For

Li^{2 +}

(Z=3), the radius of the

n

-th orbit is:

r = a_{0} \frac{n ^{2}}{Z} = a_{0} \frac{9}{3} = 3 a_{0}

Therefore:

λ = \frac{2 π \cdot 3 a _{0}}{3} = 2 π a_{0}

Hence

p = 2

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer

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