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*Simulator*

**Previous Years JEE Advanced Questions**

**Paragraph for Question Nos. 1 to 3**

The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition.

*Simulator*

**Previous Years JEE Advanced Questions**

**Paragraph for Question Nos. 1 to 3**

The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition.

$(\mathrm{A}) \frac{1}{\mathrm{n}^{2}}\left(\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}\right)$

$(\mathrm{B}) \frac{1}{\mathrm{n}}\left(\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}\right)$

$(\mathrm{C}) \mathrm{n}\left(\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}\right)$

$(\mathrm{D}) \mathrm{n}^{2}\left(\frac{\mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}\right)$

**[JEE 2010]**

**Sol.**(D)

$\mathrm{E}_{\mathrm{n}}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{(\mathrm{l} \omega)^{2}}{2 \mathrm{l}}=\frac{(\mathrm{nh} / 2 \pi)^{2}}{2 \mathrm{l}}=\frac{\mathrm{n}^{2} \mathrm{h}^{2}}{8 \pi^{2} \mathrm{I}}$

(A) $2.76 \times 10^{-46} \mathrm{kg} \mathrm{m}^{2}$

(B) $1.87 \times 10^{-46} \mathrm{kg} \mathrm{m}^{2}$

(C) $4.67 \times 10^{-47} \mathrm{kg} \mathrm{m}^{2}$

(D) $1.17 \times 10^{-47} \mathrm{kg} \mathrm{m}^{2}$

**[JEE 2010]**

**Sol.**(B)

(A) $2.4 \times 10^{-10} \mathrm{m}$

(B) $1.9 \times 10^{-10} \mathrm{m}$

(C) $1.3 \times 10^{-10} \mathrm{m}$

(D) $4.4 \times 10^{-11} \mathrm{m}$

**[JEE 2010]**

**Sol.**(C)

Moment of inertia of CO molecule about centre of mass : $\mathrm{I}=\mu \mathrm{r}^{2}$ where $\mu=\frac{\mathrm{m}_{1} \mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

**[JEE 2011]**

**Sol.**(A)

(A) $\frac{9}{32 \mathrm{R}}$

(B) $\frac{9}{16 \mathrm{R}}$

(C) $\frac{9}{\text { sR }}$

(D) $\frac{4}{3 \mathrm{R}}$

**[JEE Advanced-2013]**

**Sol.**(A,C)

for $\mathrm{n}=2$ to $\mathrm{n}=1 \quad \frac{1}{\lambda}=4 \mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right] \Rightarrow \lambda=\frac{3}{\mathrm{R}}$

**[JEE Advanced-2015]**

**Sol.**2

$\mathrm{E}_{\mathrm{Ph}}=\frac{\mathrm{hc}}{\lambda}=\frac{1242}{90}=13.8 \mathrm{eV}$

$\mathrm{E}_{\mathrm{Ph}}=\Delta \mathrm{E}+(\mathrm{K.E.})$

$13.8=\Delta \mathrm{E}+10.4$

$\Delta E=3.4 \mathrm{eV}$

so electron initially was in n = 2

(A) Relative change in the radii of two consecutive orbitals does not depend on Z

(B) Relative change in the radii of two consecutive oribitals varies as 1/n

(C) Relative change in the energy of two consecutive orbitals varies as $1 / \mathrm{n}^{3}$

(D) Relative change in the angular momenta of two consecutive orbitals varies as 1/n

**[JEE Advanced-2016]**

**Sol.**(A,B,D)

As radius $\mathrm{r} \propto \frac{\mathrm{n}^{2}}{\mathrm{z}}$

as energy $\mathrm{E} \propto \frac{\mathrm{z}^{2}}{\mathrm{n}^{2}}$

hc/e = $1.237 \times 10^{-6}$ eV m and the ground state energy of hydrogen atom as –13.6 eV, the number of lines present in the emission spectrum is ?

**[JEE Advanced-2016]**

**Sol.**6

**[JEE Advanced-2017]**

**Sol.**5