1
MCQ (Single Correct Answer)

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
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
3
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 + $${B \over {\lambda _n^2}}$$
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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