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91.
How many orbitals in total are associated with 4th energy level ?
A
4
B
9
C
16
D
7
Answer :
16
No. of orbitals $$ = {n^2},$$ if $$n = 4$$
No. of orbitals = 16
92.
The orbital angular momentum for an electron revolving in an orbit is given by $$\sqrt {l\left( {l + 1} \right)} .\frac{h}{{2\pi }}.$$ This momentum for an $$s-$$electron will be given by
A
$${\text{Zero}}$$
B
$$\frac{h}{{2\pi }}$$
C
$$\sqrt 2 .\frac{h}{{2\pi }}$$
D
$$ + \frac{1}{2}.\frac{h}{{2\pi }}$$
Answer :
$${\text{Zero}}$$
For s-electron, $$l = 0$$
∴ Orbital angular momentum = $$\sqrt {0\left( {0 + 1} \right)} \frac{h}{{2\pi }} = 0$$
93.
The energy absorbed by each molecule $$\left( {{A_2}} \right)$$ of a substance is $$4.4 \times {10^{ - 19}}J$$ and bond energy per molecule is $$4.0 \times {10^{ - 19}}J.$$ The kinetic energy of the molecule per atom will be
A
$$2.0 \times {10^{ - 20}}J$$
B
$$2.2 \times {10^{ - 19}}J$$
C
$$2.0 \times {10^{ - 19}}J$$
D
$$4.0 \times {10^{ - 20}}J$$
Answer :
$$2.0 \times {10^{ - 20}}J$$
Kinetic energy $$(KE)$$ of molecule = energy absorbed by molecule $$ - $$ bond energy per molecule
$$\eqalign{
& = \left( {4.4 \times {{10}^{ - 19}}} \right) - \left( {4.0 \times {{10}^{ - 19}}} \right)J \cr
& = 0.4 \times {10^{ - 19}}J \cr
& KE\,\,\,{\text{per atom}} = \frac{{0.4 \times {{10}^{ - 19}}}}{2}J \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2.0 \times {10^{ - 20}}J \cr} $$
94.
When electronic transition occurs from higher energy state to lower energy state with energy difference equal to $$\Delta E$$ electron volts, the wavelength of the line emitted is approximately equal to
A
$$\frac{{12395}}{{\Delta E}} \times {10^{ - 10}}m$$
B
$$\frac{{12395}}{{\Delta E}} \times {10^{10}}m$$
C
$$\frac{{12395}}{{\Delta E}} \times {10^{ - 10}}cm$$
D
$$\frac{{12395}}{{\Delta E}} \times {10^{10}}cm$$
(i) 4 $$p$$ (ii) 4 $$s$$ (iii) 3 $$d$$ (iv) 3 $$p$$
According to Bohr Bury's $$\left( {n + \ell } \right)$$
rule, increasing order of energy (iv) < (ii)< (iii) < (i). Note : If the two orbitals have same value of $$\left( {n + \ell } \right)$$ then the orbital with lower value of n will be filled first.
98.
Uncertainty in the position of an electron $$\left( {mass = 9.1 \times {{10}^{ - 31}}kg} \right)$$ moving with a velocity $$300\,m{s^{ - 1}},$$ accurate upto $$0.001\% $$ will be $$\left( {h = 6.63 \times {{10}^{ - 34}}Js} \right)$$