1
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 9th April Morning Slot

Imagine that a reactor converts all given mass into energy and that it operates at a power level of 109 watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is 3×108 m/s)
A
0.96 gm
B
0.8 gm
C
4 $\times$ 10$-$2 gm
D
6.6 $\times$ 10$-$5 gm
2
MCQ (Single Correct Answer)

### JEE Main 2018 (Offline)

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let ${\lambda _n}$, ${\lambda _g}$ be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let ${\Lambda _n}$ be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)
A
${\Lambda _n} \approx A + {B \over {\lambda _n^2}}$
B
${\Lambda _n} \approx A + B{\lambda _n}$
C
$\Lambda _n^2 \approx A + B\lambda _n^2$
D
$\Lambda _n^2 \approx \lambda$

## Explanation

We know,

Wavelength of emitted photon from n2 state to n1 state is

${1 \over \lambda }$  =  RZ2 $\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

Here electron comes from nth state to ground state (n = 1),

then the wavelength of photon is ,

${1 \over {{\Lambda _n}}}$  =  RZ2 $\left( {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right)$

$\Rightarrow $$\,\,\, \Lambda n = {1 \over {R{Z^2}}}{\left( {1 - {1 \over {{n^2}}}} \right)^{ - 1}} As n is very large, so using binomial theorem \Lambda n = {1 \over {R{Z^2}}}\left( {1 + {1 \over {{n^2}}}} \right) \Rightarrow$$\,\,\,$ $\Lambda$n =  ${1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{n^2}}}} \right)$

We know,

$\lambda$n =  ${{2\pi r} \over n}$

=  2$\pi$ ${\left( {{{{n^2}{h^2}} \over {4{\pi ^2}mZ{C^2}}}} \right)\times{{1 \over n}}}$

$\therefore\,\,\,$ $\lambda$n $\propto$ n

$\Rightarrow $$\,\,\, n = K \lambda n \therefore\,\,\, \Lambda n = {1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{{\left( {K\,{\lambda _n}} \right)}^2}}}} \right) Let A = {1 \over {R{Z^2}}} and B = {1 \over {{K^2}R{Z^2}}} \Rightarrow$$\,\,\,$ $\Lambda$n  =  A + ${B \over {\lambda _n^2}}$
3
MCQ (Single Correct Answer)

### JEE Main 2018 (Offline)

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :
A
(0, 1)
B
(0.89, 0.28)
C
(0.28, 0.89)
D
(0, 0)

## Explanation

Applying conservation of momentum :

mv + 0 = mv1 + 2mv2

$\Rightarrow $$\,\,\, v = v1 + 2v2 . . . . . (1) As collision is elastic, So, coefficient of restitution, e = 1 \therefore\,\,\, e = 1 = {{velocity\,\,of\,\,separation} \over {Velocity\,\,of\,\,approach}} \Rightarrow$$\,\,\,$ 1 = ${{{v_2} - {v_1}} \over {v - 0}}$

$\Rightarrow $$\,\,\, v = v2 - v1 . . . . .(2) Add (1) and (2), 2v = 3v2 \Rightarrow$$\,\,\,$ v2 = ${{2v} \over 3}$

put value of v2 in equation (1),

v1 = v $-$ 2v2

= v $-$ ${{4v} \over 3}$

= $-$ ${v \over 3}$

$\therefore\,\,\,$ Fractional loss of energy of neutron.

Pd = ${{{k_i} - {k_f}} \over {{k_i}}}$

= ${{{1 \over 2}m{v^2} - {1 \over 2}mv_1^2} \over {{1 \over 2}m{v^2}}}$

= ${{{v^2} - {{{v^2}} \over 9}} \over {{v^2}}}$

= ${8 \over 9}$

= 0.89

Applying momentum of conservation,

mv + 0 = mv1 + 12mv2

$\Rightarrow $$\,\,\, v = v1 + 12v2 . . . . . (3) Here also e = 1 \therefore\,\,\, e = 1 = {{{v_2} - v{}_1} \over {v - 0}} \Rightarrow$$\,\,\,$ v = v2 $-$ v1 . . . . . . (4)

adding (3) and (4), we get

2v = 13v2

$\Rightarrow $$\,\,\, v2 = {{2v} \over {13}} put this v2 in equation (3), we get v1 = v - 12 \times {{2v} \over {13}} = - {{11v} \over {13}} \therefore\,\,\, Frictional loss pc = {{{1 \over 2}m{v^2} - {1 \over 2}m{{\left( {{{11} \over {13}}v} \right)}^2}} \over {{1 \over 2}m{v^2}}} = {{48} \over {169}} = 0.28 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Offline) If the series limit frequency of the Lyman series is {\nu _L}, then the series limit frequency of the Pfund series is: A {\nu _L}/25 B 25{\nu _L} C 16{\nu _L} D {\nu _L}/16 ## Explanation Note : (1) In Lyman Series, transition happens in n = 1 state from n = 2, 3, . . . . . \propto (2) In Balmer Series, transition happens in n = 2 state from n = 3, 4, . . . . . \propto (3) In Paschen Series, transition happens in n = 3 state from n = 4, 5, . . . . . \propto (4) In Bracktt Series, transition happens in n = 4 state from n = 5, 6 . . . . . . \propto (5) In Pfund Series, transition happens in n = 5 state from n = 6, 7, . . . . \propto We know, {1 \over \lambda } = RZ2 \left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right) Series limit means transition happens from n = \propto to n = 1, for Lyman Series. In series limit for Lyman series, {1 \over {{\lambda _L}}} = RZ2 \left( {{1 \over {{1^2}}} - {1 \over \propto }} \right) \Rightarrow$$\,\,\,$ ${1 \over {{\lambda _L}}}$ = RZ2

We know,

E = ${{hc} \over \lambda }$ = h$\gamma$

$\Rightarrow $$\,\,\, \gamma = {c \over \lambda } So, frequency in Lyman Series, \gamma L = {c \over {{\lambda _L}}} = c \times RZ2 In Pfund series, n2 = \propto and n1 = 5 \therefore\,\,\, {1 \over {{\lambda _P}}} = RZ2\left( {{1 \over {{5^2}}} - {1 \over {{ \propto ^2}}}} \right) \Rightarrow$$\,\,\,$ ${1 \over {{\lambda _P}}}$ = ${{R{Z^2}} \over {25}}$

$\therefore\,\,\,$ ${\gamma _P}$   =   ${c \over {{\lambda _P}}}$ = c $\times$ ${{R{Z^2}} \over {25}}$

$\therefore\,\,\,$ $\gamma$P = ${{cRZ{}^2} \over {25}}$ = ${{{\gamma _L}} \over {25}}$    [as   $\gamma$L = cRZ2]

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