1
MCQ (Single Correct Answer)

AIPMT 2006

Given : The mass of electron is 9.11 $$ \times $$ 10$$-$$31 kg, Planck constant is 6.626 $$ \times $$ 10$$-$$34 J s, the uncertainty involved in the measurement of velocity within a distance of 0.1 $$\mathop A\limits^ \circ $$ is
A
5.79 $$ \times $$ 105 m s$$-$$1
B
5.79 $$ \times $$ 106 m s$$-$$1
C
5.79 $$ \times $$ 107 m s$$-$$1
D
5.79 $$ \times $$ 108 m s$$-$$1

Explanation

$$\Delta x.m\Delta v = h/4\pi $$

0.1 $$ \times $$ 10-10 $$ \times $$ 9.11 $$ \times $$ 10-31 $$ \times $$ $$\Delta v$$ = $${{6.626 \times {{10}^{ - 34}}} \over {4 \times 3.143}}$$

$$ \therefore $$ $$\Delta v$$ = $${{6.626 \times {{10}^{ - 34}}} \over {4 \times 3.143 \times 0.1 \times {{10}^{ - 10}} \times 9.11 \times {{10}^{ - 31}}}}$$

= 5.79 $$ \times $$ 106 m s-1
2
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.
3
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
4
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

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