1
MCQ (Single Correct Answer)

AIPMT 2006

The orientation of an atomic orbital is governed by
A
principal quantum number
B
azimuthal quantum number
C
spin quantum number
D
magnetic quantum number.

Explanation

Magnetic quantum number describes the orientation.
2
MCQ (Single Correct Answer)

AIPMT 2005

The energy of second Bohr orbit of the hydrogen atom is $$-$$328 kJ mol$$-$$1; hence the energy of fourth Bohr orbit would be
A
$$-$$ 41 kJ mol$$-$$1
B
$$-$$82 kJ mol$$-$$1
C
$$-$$164 kJ mol$$-$$1
D
$$-$$1312 kJ mol$$-$$1

Explanation

En = -K$${\left( {{Z \over n}} \right)^2}$$

Z = 1; n = 2

E2 = $${ {{-K \times 1 \over 4}}}$$$$ \Rightarrow $$ E2 = -328 kJ mol-1;

K = 4 $$ \times $$ 328

E4 = $${ {{-K \times 1 \over 16}}}$$ $$ \Rightarrow $$ E4 = -4 $$ \times $$ 328 $${ {{ 1 \over 16}}}$$

= -82 kJ mol-1
3
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
4
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

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