### JEE Mains Previous Years Questions with Solutions

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1

### 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

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

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

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

$\cong 10{}^{ - 33}$
2

### AIEEE 2003

In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen
A
5 $\to$ 2
B
4 $\to$ 1
C
2 $\to$ 5
D
3 $\to$ 2

## Explanation

In Bohr series of lines of hydrogen spectrum :- Red end means end of visible region which is balmer series of lines so n will be 2

Now for third lines with respect to Balmer series means n = (2 + 3) = 5th line

$\therefore$ n1 = 2 and n2 = 5

For 2 $\to$ 5 electron will need energy to jump as 2 > 5 so no spectrum will release instead energy will be absorbed.

$\therefore$ 5 $\to$ 2
3

### AIEEE 2003

Which one of the following groupings represents a collection of isoelectronic species? (At. nos. : Cs : 55, Br : 35)
A
N3-, F-, Na+
B
Be, Al3+, Cl-
C
Ca2+, Cs+, Br
D
Na+, Ca2+, Mg2+

## Explanation

Isoelectronic means those species whose electron number are same.

Species Atomic Number Electron Number
N3- 7 7 + 3 = 10
F- 9 9 + 1 = 10
Na+ 11 11 - 1 = 10
Be 4 4
Al3+ 13 13 - 3 = 10
Cl- 17 17 + 1 = 8
Ca2+ 20 20 - 2 = 18
Cs+ 55 55 - 1 = 54
Br 35 35
Mg2+ 12 12 -2 = 10

$\therefore$ N3-, F-, Na+
4

### AIEEE 2003

The orbital angular momentum for an electron revolving in an orbit is given by $\sqrt {l(l + 1)} {h \over {2\pi }}$. This momentum for an s-electron will be given by
A
zero
B
${h \over {2\pi }}$
C
$\sqrt 2 {h \over {2\pi }}$
D
$+ {1 \over 2}{h \over {2\pi }}$

## Explanation

For s-electron l = 0

$\therefore$ $\sqrt {l(l + 1)} {h \over {2\pi }}$

= $\sqrt {0(0 + 1)} {h \over {2\pi }}$

= $\sqrt 0 {h \over {2\pi }}$ = 0 (zero)