1
MCQ (Single Correct Answer)

AIPMT 2004

The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in hydrogen atom will be (Given ionization energy of H = 2.18 $$ \times $$ 10$$-$$18 J atom$$-$$1 and h = 6.625 $$ \times $$ 10$$-$$34 J s)
A
1.54 $$ \times $$ 1015 s$$-$$1
B
1.03 $$ \times $$ 1015 s$$-$$1
C
3.08 $$ \times $$ 1015 s$$-$$1
D
2.00 $$ \times $$ 1015 s$$-$$1

Explanation

E = h$$v$$ or $$v$$ = E/h

For H atom, E = $${{ - 21.76 \times {{10}^{ - 19}}} \over {{n^2}}}J\,at{m^{ - 1}}$$

$$\Delta E = - 21.76 \times {10^{ - 19}}\left( {{1 \over {{4^2}}} - {1 \over {{1^2}}}} \right)$$

= 20.40 $$ \times $$ 10-19 J atm-1

$$v$$ = $${{20.40 \times {{10}^{ - 19}}} \over {6.626 \times {{10}^{ - 34}}}}$$ = 3.079 $$ \times $$ 1015 s-1
2
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
3
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
4
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)

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