JEE Main Atom PYQ

26. (JEE Main 2019 (Online) 12th April Evening Slot)

Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 $\overset{o}{A}$ ). The de-Broglie wavelength of this electron is :

A. 6.6 $\overset{o}{A}$

B. 3.5 $\overset{o}{A}$

C. 9.7 $\overset{o}{A}$

D. 12.9 $\overset{o}{A}$

The Correct Answer is Option (C)

For second excited state n = 3

$m v r = \frac{3 h}{2 π}$ ........(1)

$m v = \frac{h}{λ}$ .........(2)

Dividing (1) by (2), we get

$r = \frac{3 λ}{2 π}$

$\Rightarrow$ $λ = \frac{2 π r}{3}$ = $\frac{2 \times 3.14 \times 4.65}{3}$ = 9.7 $\overset{o}{A}$

27. (JEE Main 2019 (Online) 9th April Evening Slot)

A He⁺ ion is in its first excited state. Its ionization energy is :

A. 13.60 eV

B. 6.04 eV

C. 48.36 eV

D. 54.40 eV

The correct answer is option (A)

$T . E . = - (13.6) (\frac{Z^{2}}{n^{2}}) e V$

z = n = 2

$\Rightarrow$ Ionisation energy = –T.E. = 13.6 eV

28. (JEE Main 2019 (Online) 12th January Morning Slot)

A particle of mass m moves in a circular orbit in a central potential field U(r) = $\frac{1}{2}$ kr². If Bohr 's quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as :

A. r_n $\propto$ $\sqrt{n}$ , E_n $\propto$ n

B. r_n $\propto$ $\sqrt{n}$ , E_n $\propto$ $\frac{1}{n}$

C. r_n $\propto$ n, E_n $\propto$ n

D. --> $\propto$ n², E_n $\propto$ $\frac{1}{n^{2}}$

The correct answer is option (C)

Force due to this field, F = $- \frac{\partial U}{\partial r}$

F = $- \frac{\partial}{\partial r} (\frac{1}{2} k r^{2})$ = -kr

For circular orbit, $\frac{m v^{2}}{r}$ = -kr

$\Rightarrow$ v $\propto$ r ..... (1)

Fron Bohr’s quantization condition

mvr = $\frac{n h}{2 π}$ .....(2)

From (1) and (2),

$r_{n}$ $\propto$ $n^{\frac{1}{2}}$

Given, U(r) = $\frac{1}{2}$ kr²

$\Rightarrow$ E_n = $- \frac{1}{2} U (r)$ = $- \frac{1}{4} k r^{2}$

$\Rightarrow$ E_n $\propto$ n

29. (JEE Main 2018 (Online) 15th April Evening Slot)

Muon ( $μ$ ^$-$) is a negatively charged (|q| = |e|) particle with a mass m_$μ$ = 200 m_e, where m_e is the mass of the electron and e is the electronic charge. If $μ$ ^$-$ is bond to a proton to form a hydrogen like atom, identify the correct statements.
(A)   Radis of the muonic orbit is 200 times smaller than that of the electron.
(B)   The speed of the $μ$ ^$-$ in the n^th orbit is $\frac{1}{200}$ times that of the electron in the n^th orbit.
(C)   The ionization energy of muonic atom is 200 timesmore than of an hydroen atom.
(D)   The momentum of the muon in the n^th orbit is 200 times more than that of the electron.

A. (A), (B), (D)

B. (A), (C), (D)

C. (B), (D)

D. (C), (D)

The correct answer is option (B)

(i) Radius of the orbit is given by

$r = \frac{h^{2}}{4 π m e^{2}} \times \frac{n^{2}}{Z}$

Only, m_$μ$ = 200 m_e rest are same. So, statement (A) is correct: The radius of muonic orbit is 200 times smaller than that of the electron.

(ii) Velocity of particle in an orbit is given by

$v = \frac{n h}{2 π m r} = \frac{n h \times 4 π m e^{2} \times Z}{2 π m \times h^{2} \times n^{2}} = \frac{2 e^{2} Z}{h n}$

Therefore, speed does not change as it does not depend on mass. So, statement (B) is incorrect.

(iii) Ionisation energy is given by

$Δ E_{n} = - \frac{2 π^{2} m e^{4} Z^{2}}{h^{2} {(4 π ε_{0})}^{2}} \times (\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}})$

Only, m_$μ$ = 200 m_e rest are same. So, statement (C) is correct: The ionisation energy of muonic atom is 200 times more than that of an hydrogen atom.

(iv) Since Momentum $\propto$ Energy. So, statement (D) is correct: The momentum of the muon in he nth orbit is 200 times more than that of the electron.

30. (JEE Main 2018 (Online) 15th April Morning Slot)

The energy required to remove the electron from a singly ionized Helium atom is $2.2$ times the energies required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is :

A. $20$ $e V$

B. $34$ $e V$

C. $79$ $e V$

D. $109$ $e V$

The correct answer is option (C)

Energy required to remove e^$-$ from singly ionized Helium atom

E₁ = $\frac{13.6 z^{2}}{n^{2}}$ = $\frac{13.6 \times 2^{2}}{1^{2}}$ = 54.4 eV.

Let, E₂ = energy required to remove e^$-$from He $-$ atom

$∴$ According to q vertion.

E₁ = 2.2 E₂

$∴$ E₂ = $\frac{54.4}{2.2}$ = 24.72 eV

$∴$ Total energy required to ionize Helium atom completely

= (54.4 + 24.72) eV

= 79.12 eV