1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 9th April Morning Slot

A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :
A
2
B
3
C
4
D
5

Explanation

Energy released when hydrogen atom makes transition from n = 2 to n = 1 is,

E1 = 13.6 $$ \times $$ $$\left( {{1 \over {{1^2}}} - {1 \over {{z^2}}}} \right)$$

= $${3 \over 4} \times 13.6\,\,\,$$ eV

Energy required to remove a electron from nth excited state of doubly ionized lithium,

E2 = $${{13.6{z^2}} \over {{n^2}}}$$ = $${{13.6 \times {3^2}} \over {{n^2}}}$$ eV

This energy is provided by the photon when it strike with the lithium atom.

$$ \therefore $$   E1 $$ \ge $$ E2

$$ \Rightarrow $$   $${3 \over 4} \times 13.6 \ge {{13.6 \times 9} \over {{n^2}}}$$

$$ \Rightarrow $$   n2   $$ \ge $$  3$$ \times $$4

$$ \Rightarrow $$  n   $$ \ge $$ $$\sqrt {12} $$

$$ \Rightarrow $$    n $$ \ge $$ 3.5

$$ \therefore $$  Least possible excited state = 4
2
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = $${{\lambda _1}}$$/$${{\lambda _2}}$$, is given by:
A
r = 1/3
B
r = 4/3
C
r = 2/3
D
r = 3/4

Explanation

From energy level diagram, using $$\Delta E = {{hc} \over \lambda }$$

For wavelength $${{\lambda _1}}$$, $$\Delta E =$$ –E – (–2E) = $${{hc} \over {{\lambda _1}}}$$

For wavelength $${{\lambda _2}}$$, $$\Delta E =$$ –E – (–$${{4E} \over 3}$$) = $${{hc} \over {{\lambda _2}}}$$

$$ \therefore $$ r = $${{{\lambda _1}} \over {{\lambda _2}}} = {{\left( {{{4E} \over 3} - E} \right)} \over {\left( {2E - E} \right)}}$$ = $${1 \over 3}$$
3
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :
A
$$t = {T \over {\log (1.3)}}$$
B
$$t = T\log (1.3)$$
C
$$t = {T \over 2}{{\log 2} \over {\log 1.3}}$$
D
$$t = T{{\log 1.3} \over {\log 2}}$$

Explanation

Let initially there are total N0 number of nuclei.

At time t $${{{N_B}} \over {{N_A}}}$$ = 0.3 (given)

$$ \Rightarrow $$ $${{N_B}}$$ = 0.3$${{N_A}}$$

N0 = NA + NB = NA + 0.3NA

$$ \therefore $$ NA = $${{{N_0}} \over {1.3}}$$

As we know Nt = N0 e– $$\lambda $$t

$$ \Rightarrow $$ $${{{N_0}} \over {1.3}}$$ = N0 e– $$\lambda $$t

$$ \Rightarrow $$ $${1 \over {1.3}}$$ = e– $$\lambda $$t

$$ \Rightarrow $$ ln(1.3) = $$\lambda $$t

$$ \Rightarrow $$ t = $${{\ln \left( {1.3} \right)} \over \lambda }$$

If T is half-life, then $$\lambda $$ = $${{{{\ln} } 2} \over T}$$

$$ \Rightarrow $$ t = $${{\ln \left( {1.3} \right)} \over {{{\ln 2} \over T}}}$$ = $${{\ln \left( {1.3} \right)} \over {\ln 2}}T$$
4
MCQ (Single Correct Answer)

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 nth orbit is proportional to : (n = principal quantum number)
A
$${n^{ - 4}}$$
B
$${n^{ - 5}}$$
C
n$$-$$3
D
n$$-$$2

Explanation

Magnetic field at the center of neucleus of H-atom

B = $${{{\mu _0}I} \over {2{r_n}}}$$

Radius of nth orbital,

rn = $${{{n^2}{h^2}{\varepsilon _0}} \over {m\pi Z{e^2}}}$$

$$\therefore\,\,\,$$ rn $$ \propto $$ n2

velocity of electron in nth orbital,

$$\upsilon $$n = $$\left( {{{{e^2}h} \over {2{\varepsilon _0}}}} \right){Z \over n}$$

$$\therefore\,\,\,$$ $$\upsilon $$n $$ \propto $$ n$$-$$1

I = $${q \over t}$$ = $${e \over {{{2\pi {r_n}} \over {{\upsilon _n}}}}}$$ = $${{e{\upsilon _n}} \over {2\pi {r_n}}}$$

$$\therefore\,\,\,$$ B = $${{{\mu _0}.\left( {{{e{\upsilon _n}} \over {2\pi {r_n}}}} \right)} \over {2{r_n}}}$$ = $${{{\mu _0}\,e{\upsilon _n}} \over {4\pi r_n^2}}$$

$$\therefore\,\,\,$$ B $$ \propto $$ $${{{\upsilon _n}} \over {r_n^2}}$$

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

$$ \Rightarrow $$$$\,\,\,$$ B $$ \propto $$ n $$-$$5

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