### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2004

The wavelength of the radiation emitted when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097 $\times$ 107 m-1)
A
406 nm
B
192 nm
C
91 nm
D
9.1 $\times$ 10-8 nm

## Explanation

We know Rydberg formula,

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

[ for hydrogen atom Z = 1 ]

= $1.097 \times {10^7}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$

$\therefore$ $\lambda$ = ${1 \over {1.097 \times {{10}^7}}}$

= 9.11 $\times$10-8 m

= 91.1 $\times$10-9 m

= 91.1 nm

[ as 1 nm = 10-9 m ]
2

### AIEEE 2004

Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers, l = 1 and 2 are respectively
A
16 and 4
B
12 and 5
C
12 and 4
D
16 and 5

## Explanation

Electronic configuration of Cr (Z = 24) =

1s2 2s2 2p6 3s2 3p6 4s1 3d5

For 2p6 and 3p6 , l = 1. Here in 2p and 3p orbital total 6 + 6 = 12 electrons present.

For 3d5 , l = 2. Here in 3d orbital total 5 electrons present.
3

### AIEEE 2004

Which of the following sets of quantum numbers is correct for an electron in 4f orbital?
A
n = 4, l = 3, m = +1, s = + 1/2
B
n = 4, l = 4, m = -4, s = - 1/2
C
n = 4, l = 3, m = +4, s = + 1/2
D
n = 3, l = 2, m = -2, s = + 1/2

## Explanation

For 4f orbital,

n = 4 and l = 3

Note : ( for s orbital l = 0, for p orbital l = 1, for d orbital l = 2, for f orbital l = 3)

Value of m = + l to - l

Here m = +3 to -3

Value of s = $+ {1 \over 2}$ or $- {1 \over 2}$
4

### AIEEE 2003

The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 meters per second is approximately
A
10-31 meters
B
10-16 meters
C
10-25 meters
D
10-33 meters

## Explanation

As per de-Broglie wavelength of a particle is

Wavelength ($\lambda$) = ${h \over {mv}}$

Where h is plank's constant, m is mass of the particle and v is the velocity of the particle

$\therefore \lambda$ = ${{6.63 \times {{10}^{ - 34}}} \over {0.06 \times 10}}$

= ${{6.63 \times {{10}^{ - 34}} \times 10} \over {0.6 \times 10}}$

= ${{6.63 \times {{10}^{ - 33}}} \over 6}$

$\cong 10{}^{ - 33}$