1
MCQ (Single Correct Answer)

AIPMT 2014

Calculate the energy in joule corresponding to light of wavelength 45 nm. (Planck,s constant, h = 6.63 $$ \times $$ 10$$-$$34 J s, speed of light, c = 3 $$ \times $$ 108 m s$$-$$1)
A
6.67 $$ \times $$ 1015
B
6.67 $$ \times $$ 1011
C
4.42 $$ \times $$ 10$$-$$15
D
4.42 $$ \times $$ 10$$-$$18

Explanation

E = $$hc \over \lambda$$

E = $${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {45 \times {{10}^{ - 9}}}}$$

= 4.42 $$ \times $$ 10-18 J
2
MCQ (Single Correct Answer)

AIPMT 2014

What is the maximum number of orbitals that can be identified with the following quantum numbers ?
n = 3, l = 1, m1 = 0
A
1
B
2
C
3
D
4

Explanation

Only one orbital 3p has following set of quantum numbers, n = 3, l = 1 and ml = 0
3
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as (h = 6.62 $$ \times $$ 10$$-$$27 ergs, c = 3 $$ \times $$ 1010 cm s$$-$$1, NA = 6.02 $$ \times $$ 10$$-$$23 mol$$-$$1)
A
$${{1.196 \times {{10}^8}} \over \lambda }$$
B
$${{2.859 \times {{10}^5}} \over \lambda }$$
C
$${{2.859 \times {{10}^{16}}} \over \lambda }$$
D
$${{1.196 \times {{10}^{16}}} \over \lambda }$$

Explanation

E = $${{hc{N_A}} \over \lambda }$$

= $${{6.62 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}} \times 6.02 \times {{10}^{23}}} \over \lambda }$$

= $${{1.1955 \times {{10}^8}} \over \lambda }$$ = $${{1.196 \times {{10}^8}} \over \lambda }$$ ergs mol-1
4
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

The outer electronic configuration of Gd (At. No. 64) is
A
4f5 5d4 6s1
B
4f7 5d1 6s2
C
4f3 5d5 6s2
D
4f4 5d5 6s1

Explanation

Gd[64] = [Xe]4f7 5d1 6s2

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