JEE Main Atom PYQ

31. (JEE Main 2017 (Online) 9th April Morning Slot)

The acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is :

A. $\frac{h^{2}}{π^{2} m^{2} r^{3}}$

B. $\frac{h^{2}}{8 π^{2} m^{2} r^{3}}$

C. $\frac{h^{2}}{4 π^{2} m^{2} r^{3}}$

D. $\frac{h^{2}}{4 π m^{2} r^{3}}$

The correct answer is option (C)

The speedy of the particle in the orbit of an atom is $v = \frac{I^{2}}{2 h ε_{0}}$

We have the radius of the first orbit is

$r = \frac{h^{2} ε_{0}}{π m e^{2}}$

$\Rightarrow ε_{0} = \frac{r π m e^{2}}{h^{2}}$

Therefore, the acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is

$\frac{v^{2}}{r} = \frac{I^{6} π m}{4 h^{2} ε_{0}^{3}} = \frac{h^{2}}{4 r^{3} π^{2} m^{2}}$

32. (JEE Main 2017 (Online) 8th April Morning Slot)

According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the n^th orbit is proportional to : (n = principal quantum number)

A. $n^{- 4}$

B. $n^{- 5}$

C. n^{$-$ 3}

D. n^{$-$ 2}

The correct answer is option (B)

Magnetic field at the center of neucleus of H-atom

B = $\frac{μ_{0} I}{2 r_{n}}$

Radius of n^th orbital,

r_n = $\frac{n^{2} h^{2} ε_{0}}{m π Z e^{2}}$

$∴$ r_n $\propto$ n²

velocity of electron in n^th orbital,

$υ$ _n = $(\frac{e^{2} h}{2 ε_{0}}) \frac{Z}{n}$

$∴$ $υ$ _n $\propto$ n^{$-$ 1}

I = $\frac{q}{t}$ = $\frac{e}{\frac{2 π r_{n}}{υ_{n}}}$ = $\frac{e υ_{n}}{2 π r_{n}}$

$∴$ B = $\frac{μ_{0} . (\frac{e υ_{n}}{2 π r_{n}})}{2 r_{n}}$ = $\frac{μ_{0} e υ_{n}}{4 π r_{n}^{2}}$

$∴$ B $\propto$ $\frac{υ_{n}}{r_{n}^{2}}$

$\Rightarrow$ B $\propto$ $\frac{n^{- 1}}{n^{4}}$

$\Rightarrow$ B $\propto$ n ^{$-$ 5}

33. ( JEE Main 2015 (Offline))

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion :

A. kinetic energy decreases, potential energy increases but total energy remains same

B. kinetic energy and total energy decrease but potential energy increases

C. its kinetic energy increases but potential energy and total energy decrease

D. kinetic energy, potential energy and total energy decrease

The correct answer is option (C)

$U = - K \frac{z e^{2}}{r}; T . E = \frac{k}{2} \frac{z e^{2}}{r}$

$K . E = \frac{k}{2} \frac{z e^{2}}{r} .$ Here $r$ decreases

34. (AIEEE 2012)

A diatomic molecule is made of two masses $m_{1}$ and $m_{2}$ which are separated by a distance $r .$ If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: ( $n$ is an integer)

A. $\frac{{(m_{1} + m_{2})}^{2} n^{2} h^{2}}{2 m_{1}^{2} m_{2}^{2} r^{2}}$

B. $\frac{n^{2} h^{2}}{2 (m_{1} + m_{2}) r^{2}}$

C. $\frac{2 n^{2} h^{2}}{(m_{1} + m_{2}) r^{2}}$

D. $\frac{(m_{1} + m_{2}) n^{2} h^{2}}{2 m_{1} m_{2} r^{2}}$

The correct answer is option (D)

The energy of the system of two atoms of diatomic

molecule $E = \frac{1}{2} I ω$

where $I =$ moment of inertia

$ω =$ Angular velocity $= \frac{L}{I} .$

$L =$ Angular momentum

$I = \frac{1}{2} (m_{1} {r_{1}}^{2} + m^{2} {r_{2}}^{2})$

Thus, $E = \frac{1}{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2}) ω^{2} . . . (i)$

$E = \frac{1}{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2}) \frac{L^{2}}{I^{2}}$

$L = n \frac{n h}{2 n}$ (According Bohr's Hypothesis)

$E = \frac{1}{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2}) \frac{L^{2}}{{(m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2})}^{2}}$

$E = \frac{1}{2} \frac{L^{2}}{(m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2})}$

$= \frac{n^{2} h^{2}}{8 π^{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2})}$

$E = \frac{(m_{1} + m_{2}) n^{2} h^{2}}{8 π^{2} r^{2} m_{1} m_{2}}$

$[$ as $r_{1} = \frac{m_{2} r}{m_{1} + m_{2}}; r_{2} = \frac{m_{2} r}{m_{1} + m_{2}}]$

35. (AIEEE 2012)

A diatomic molecule is made of two masses $m_{1}$ and $m_{2}$ which are separated by a distance $r .$ If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: ( $n$ is an integer)

A. $\frac{{(m_{1} + m_{2})}^{2} n^{2} h^{2}}{2 m_{1}^{2} m_{2}^{2} r^{2}}$

B. $\frac{n^{2} h^{2}}{2 (m_{1} + m_{2}) r^{2}}$

C. $\frac{2 n^{2} h^{2}}{(m_{1} + m_{2}) r^{2}}$

D. $\frac{(m_{1} + m_{2}) n^{2} h^{2}}{2 m_{1} m_{2} r^{2}}$

The correct answer is option (D)

The energy of the system of two atoms of diatomic

molecule $E = \frac{1}{2} I ω$

where $I =$ moment of inertia

$ω =$ Angular velocity $= \frac{L}{I} .$

$L =$ Angular momentum

$I = \frac{1}{2} (m_{1} {r_{1}}^{2} + m^{2} {r_{2}}^{2})$

Thus, $E = \frac{1}{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2}) ω^{2} . . . (i)$

$E = \frac{1}{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2}) \frac{L^{2}}{I^{2}}$

$L = n \frac{n h}{2 n}$ (According Bohr's Hypothesis)

$E = \frac{1}{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2}) \frac{L^{2}}{{(m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2})}^{2}}$

$E = \frac{1}{2} \frac{L^{2}}{(m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2})}$

$= \frac{n^{2} h^{2}}{8 π^{2} (m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2})}$

$E = \frac{(m_{1} + m_{2}) n^{2} h^{2}}{8 π^{2} r^{2} m_{1} m_{2}}$

$[$ as $r_{1} = \frac{m_{2} r}{m_{1} + m_{2}}; r_{2} = \frac{m_{2} r}{m_{1} + m_{2}}]$