### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h / $\lambda$ (where $\lambda$ is wavelength associated with electron wave) is given by :
A
2meV
B
$\sqrt {meV}$
C
$\sqrt {2meV}$
D
meV

## Explanation

We know, Kinetic Energy (KE) = ${1 \over 2}m{v^2}$

$\therefore$ mv2 = 2KE

$\Rightarrow$ m2v2 = 2mKE

$\Rightarrow$ mv = $\sqrt {2m{K_E}}$

For an electron of charge 'e' which is passes through 'V' volt, kinetic energy of electron will be

KE = eV

$\therefore$ m = $\sqrt {2meV}$

We know de-Broglie wavelength for an electron$\left( \lambda \right)$ = ${h \over p} = {h \over {mv}}$

$\therefore$ ${h \over \lambda }$ = mv = $\sqrt {2meV}$
2

### JEE Main 2015 (Offline)

Which of the following is the energy of a possible excited state of hydrogen?
A
–6.8 eV
B
–3.4 eV
C
+6.8 eV
D
+13.6 eV

## Explanation

Energy(En) = $- {{13.6{Z^2}} \over {{n^2}}}eV$

where n = 1, 2, 3, 4, ......

For hydrogen Z = 1 and excited state starts from n = 2

So putting n = 2

E2 = $- {{13.6} \over {{2^2}}}eV$ = -3.4 eV
3

### JEE Main 2014 (Offline)

The correct set of four quantum numbers for the valence elections of rubidium atom (Z= 37) is:
A
5, 1, 1, + 1/2
B
5, 1, 0, + 1/2
C
5, 0, 0, + 1/2
D
5, 0, 1, + 1/2

## Explanation

The electronic configuration of Rubidium (Rb = 37):
1s22s22p63s23p63d104s24p65s1

As you can see last electron or valence electron enterin 5s subshell

So, the quantum numbers are n = 5, $l$ = 0, m = 0, s = $\pm {1 \over 2}$
4

### JEE Main 2013 (Offline)

Energy of an electron is given by $E = - 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)$. Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be
(h = 6.62 × 10−34 Js and c = 3.0 × 108 ms−1)
A
2.816 × 10−7 m
B
6.500 × 10−7 m
C
8.500 × 10−7 m
D
1.214 × 10−7 m

## Explanation

Given $E$ = $- 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)$
= $- 2.178 \times {10^{ - 8}}{Z^2}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$
Electron in an hydrogen atom exited from level n = 1 to n = 2. So n2 = 2 and n1 = 1.
And for H, Z = 1
$\therefore$ $E = - 2.178 \times {10^{ - 8}}\left( {{1 \over {{{\left( 2 \right)}^2}}} - {1 \over {{{\left( 1 \right)}^2}}}} \right)$
= 1.6335 $\times$ 10-18 J

We know $E$ = ${{hc} \over \lambda }$
$\therefore$ ${{hc} \over \lambda }$ = 1.6335 $\times$ 10-18
then $\lambda = {{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6335 \times {{10}^{ - 18}}}}$
$= 1.214 \times {10^{ - 7}}m$