An electron in an excited state of ion has angular momentum . The de Broglie wavelength of the electron in this state is (where is the Bohr radius). The value of is [2015]

Question

An electron in an excited state of ${Li}^{2 +}$ ion has angular momentum $\frac{3 h}{2 π}$ . 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 $p$ is [2015]

Smart Achievers · Accepted Answer

(2)

Angular momentum, $L = m v r = \frac{n h}{2 π} = \frac{3 h}{2 π}$ $p$

$∵$ $λ = \frac{h}{p} = \frac{h}{m v} = \frac{h r}{m v r} \Rightarrow m v r = \frac{h r}{λ}$ $∴$ $\frac{h r}{λ} = \frac{3 h}{2 π}$

$∴$ $λ = \frac{2 π r}{3} = \frac{2}{3} π [a_{0} \frac{n^{2}}{z}]$ $[∵ r = a_{0} \frac{n^{2}}{z}]$

$∴$ $λ = \frac{2}{3} π a_{0} [\frac{3 \times 3}{3}] = 2 π a_{0}$ $[∵ n = 3, z = 3]$

$∴$ $p = 2$

Solution

Q.
An electron in an excited state of ${Li}^{2 +}$ ion has angular momentum $\frac{3 h}{2 π}$ . 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 $p$ is [2015]

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Solution

Q. An electron in an excited state of Li2+ ion has angular momentum 3h2π. The de Broglie wavelength of the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is [2015]

Ans. (2) Angular momentum, L=mvr=nh2π=3h2π⇒n=3 ∵ λ=hp=hmv=hrmvr⇒mvr=hrλ ∴ hrλ=3h2π ∴ λ=2πr3=23π[a0n2z] [∵r=a0n2z] ∴ λ=23πa0[3×33]=2πa0 [∵n=3, z=3] ∴ p=2

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Q.
An electron in an excited state of ${Li}^{2 +}$ ion has angular momentum $\frac{3 h}{2 π}$ . 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 $p$ is [2015]