1

### 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

### 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

### 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

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

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