1

### JEE Main 2017 (Online) 8th April Morning Slot

Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)
A
30.2 MeV
B
32.4 MeV
C
23.6 MeV
D
25.8 MeV

## Explanation

1H2 + 1H2  $\to$  2He4

No. of proton in one dueteron = 2

$\therefore\,\,\,$ Total protons in two dueterons = 2 $\times$ 2 = 4

$\therefore\,\,\,$ Binding energy of two dueteron

= 1.1 $\times$ 4 = 4 : 4 MeV

In (2He4) no of protons = 4

$\therefore\,\,\,$ Binding energy of (2He4) nuclei = 4 $\times$ 7 = 28 Mev

Energy released in this process = 28 $-$ 4.4 = 23.6 MeV
2

### 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
${{{h^2}} \over {{4\pi }{m^2}{r^3}}}$
3

### 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
4

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