1

### 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
${{{h^2}} \over {{\pi ^2}{m^2}{r^3}}}$
B
${{{h^2}} \over {{8\pi ^2}{m^2}{r^3}}}$
C
${{{h^2}} \over {{4\pi ^2}{m^2}{r^3}}}$
D
${{{h^2}} \over {{4\pi }{m^2}{r^3}}}$
2

### 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
3

### 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 + ${B \over {\lambda _n^2}}$
4

### 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]