1
MCQ (Single Correct Answer)

AIPMT 2003

The velocity of Planck's constant is 6.63 $$ \times $$ 10$$-$$34 J s.

The velocity of light is 3.0 $$ \times $$ 108 m s$$-$$1. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8 $$ \times $$ 1015 s$$-$$1 ?
A
2 $$ \times $$ 10$$-$$25
B
5 $$ \times $$ 10$$-$$18
C
$$4 \times {10^1}$$
D
3 $$ \times $$ 107

Explanation

$$v$$ = c/$$\lambda $$

$$\lambda $$ = $${c \over v}$$ = $${{3 \times {{10}^8}} \over {8 \times {{10}^{15}}}}$$ = 37.5 $$ \times $$ 10-9 m

= 37.5 nm $$ \approx $$ $$4 \times {10^1}$$ nm
2
MCQ (Single Correct Answer)

AIPMT 2002

In hydrogen atom, energy of first excited state is $$-$$ 3.4 eV. Then find out K.E. of same orbit of hydrogen atom
A
+3.4 eV
B
+6.8 eV
C
$$-$$ 13.6 eV
D
+13.6 eV

Explanation

K.E = 1/2 mv2

= $${\left( {{{\pi {e^2}} \over {nh}}} \right)^2} \times 2m$$ [ $$ \because $$ v = $${{2\pi {e^2}} \over {nh}}$$]

$$ \therefore $$ Total energy = En = $${{ - 2{\pi ^2}m{e^4}} \over {{n^2}{h^2}}}$$

= -$${\left( {{{\pi {e^2}} \over {nh}}} \right)^2}$$ $$ \times $$ 2m = -K.E

$$ \therefore $$ K.E = - En

Energy of first excited state is -3.4 eV

$$ \therefore $$ Kinetic energy of the same orbit (n = 2) will be +3.4 ev
3
MCQ (Single Correct Answer)

AIPMT 2001

The following quantum numbers are possible for how many orbitals :

n = 3, $$l$$ = 2, m = +2 ?
A
1
B
2
C
3
D
4

Explanation

n = 3, l = 2, m = +2

It shows one of the five d-orbitals(3d)
4
MCQ (Single Correct Answer)

AIPMT 2001

Main axis of a diatomic molecule is z, molecular orbital px and py overlap to from which of the following orbitals.
A
$$\pi $$ molecular orbital
B
$$\sigma $$ molecular orbital
C
$$\delta $$ molecular orbital
D
No bond will form

Explanation

For $$\pi $$ overlap, the lobes of the atomic orbitals are perpendicularto the line joining the nuclei.

Hence only sidewise overlapping takes place.

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