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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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)

Questions Asked from Structure of Atom

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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