The radius of the orbit of an electron in a Hydrogen-like atom is , where is the Bohr radius. Its orbital angular momentum is It is given that is Planck's constant and is the Rydberg constant. The possible wavelength(s), when the atom de-excites, is (

Question

The radius of the orbit of an electron in a Hydrogen-like atom is $4.5 a_{0}$ , where $a_{0}$ is the Bohr radius. Its orbital angular momentum is $\frac{3 h}{2 π} .$ It is given that $h$ is Planck's constant and $R$ is the Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are) [2013]

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(1, 3)

According to Bohr's quantisation, angular momentum

$L = \frac{n h}{2 π} = \frac{3 h}{2 π}$ $∴$ $n = 3$

Also, $r_{n} = \frac{a_{0} n^{2}}{Z} = 4.5 a_{0}$

$∴$ $\frac{n^{2}}{Z} = 4.5$ $\Rightarrow \frac{9}{Z} = 4.5$ $\Rightarrow Z = 2$

According to Rydberg formula,

$\frac{1}{λ} = R Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] = 4 R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

For $n_{2} = 3, \to n_{1} = 1$ , $λ = \frac{9}{8 \times 4 R} = \frac{9}{32 R}$

For $n_{2} = 3, \to n_{1} = 2$ , $λ = \frac{36}{5 \times 4 R} = \frac{9}{5 R}$

For $n_{2} = 2, \to n_{1} = 1$ , $λ = \frac{4}{3 \times 4 R} = \frac{1}{3 R}$

Solution

1 $\frac{9}{32 R}$

2 $\frac{9}{16 R}$

3 $\frac{9}{5 R}$

4 $\frac{4}{3 R}$

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Solution

Q. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the Bohr radius. Its orbital angular momentum is 3h2π. It is given that h is Planck's constant and R is the Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are) [2013]

1 932R

2 916R

3 95R

4 43R

Ans. (1, 3) According to Bohr's quantisation, angular momentum L=nh2π=3h2π ∴ n=3 Also, rn=a0n2Z=4.5a0 ∴ n2Z=4.5⇒9Z=4.5⇒Z=2 According to Rydberg formula, 1λ=RZ2[1n12-1n22]=4R[1n12-1n22] For n2=3,→n1=1, λ=98×4R=932R For n2=3,→n1=2, λ=365×4R=95R For n2=2,→n1=1, λ=43×4R=13R

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1 $\frac{9}{32 R}$

2 $\frac{9}{16 R}$

3 $\frac{9}{5 R}$

4 $\frac{4}{3 R}$