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 in that orbit is 3h2π. It is given that ‘h’ is Plank constant and ‘R’ is Rydberg constant. Then one of the possible wavelength,when the atom de-excites is N32R. Here ‘N’ value is (‘N’ is integer and single digit)

L=nh2π=3h2π, n=3, rn=r0n2z=4.5 a0⇒ z= here a0=r0 1λ=RZ2 (1n12−1n22)=4R(1n12−1n22) For n2=3→n1=1, λ=932R For n2=3→n1=2, λ=95R For n2=2→n1=1 λ=13R

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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 in that orbit is $\frac{3 h}{2 π}$ . It is given that ‘h’ is Plank constant and ‘R’ is Rydberg constant. Then one of the possible wavelength,
when the atom de-excites is $\frac{N}{32 R}$ . Here ‘N’ value is (‘N’ is integer and single digit)

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Detailed Solution

$L = \frac{n h}{2 π} = \frac{3 h}{2 π}, n = 3, r_{n} = \frac{r_{0} n^{2}}{z} = 4.5 a_{0} \Rightarrow z = h e r e a_{0} = r_{0} \frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = 4 R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) F o r n_{2} = 3 \to n_{1} = 1, λ = \frac{9}{32 R} F o r n_{2} = 3 \to n_{1} = 2, λ = \frac{9}{5 R} F o r n_{2} = 2 \to n_{1} = 1 λ = \frac{1}{3 R}$