Practice Start Test

Structure of Atom

NEET Chemistry · 92 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
The orbital angular momentum of a p-electron is given as
A h2π{h \over {\sqrt 2 \pi }}
B 3h2π\sqrt 3 {h \over {2\pi }}
C 32hπ\sqrt {{3 \over 2}} {h \over \pi }
D 6h2π\sqrt 6 {h \over {2\pi }}
Correct Answer
Option A
Solution

Orbital angular momentum (m) =

l(l+1)h2π\sqrt {l(l + 1)} {h \over {2\pi }}

For p-electrons; l = 1 Thus, m =

l(l+1)h2π\sqrt {l(l + 1)} {h \over {2\pi }}

=

2h2π{ \sqrt 2 {h}} \over 2 \pi

=

h2π{h \over {\sqrt 2 \pi }}
Q32
Maximum number of electrons in a subshell with ll = 3 and n = 4 is
A 14
B 16
C 10
D 12
Correct Answer
Option A
Solution

l = 3 and n = 4 represent 4f, So total number of electrons = 2(2

ll

+1) = 14

Q33
The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is
A 5, 1, 1, + 1/2
B 6, 0, 0, +1/2
C 5, 0, 0, +1/2
D 5, 1, 0, +1/2
Correct Answer
Option C
Solution

Rb(37) = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 For 5s, n= 5, l = 0, m = 0, s = +1/2 or -1/2

Q34
According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon ?
A n = 6 to n = 1
B n = 5 to n = 4
C n = 6 to n = 5
D n = 5 to n = 3
Correct Answer
Option C
Solution

We know that

ΔE∝[1n12−1n22]\Delta E \propto \left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]

, where n 2 > n 1 ∴\therefore n = 6 and n = 5 will give least energetic photon.

Q35
The total number of atomic orbitals in fourth energy level of an atom is
A 8
B 16
C 32
D 4
Correct Answer
Option B
Solution

The total number of atomic orbitals in any energy level is n 2 . ∴\therefore 4 2 = 16

Q36
The energies E 1 and E 2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e., λ\lambda 1 and λ\lambda 2 will be
A λ\lambda 1 = λ\lambda 2
B λ\lambda 1 = 2λ\lambda 2
C λ\lambda 1 = 4λ\lambda 2
D λ\lambda 1 = 12{1 \over 2} λ\lambda 2
Correct Answer
Option B
Solution

E 1 =

hcλ1{{hc} \over {{\lambda _1}}}

and E 2 =

hcλ2{{hc} \over {{\lambda _2}}}
E1E2{{{E_1}} \over {{E_2}}}

=

hcλ1×λ2hc{{hc} \over {{\lambda _1}}} \times {{{\lambda _2}} \over {hc}}

=

λ2λ1{{{\lambda _2}} \over {{\lambda _1}}}

⇒\Rightarrow

255025 \over 50

=

λ2λ1{{{\lambda _2}} \over {{\lambda _1}}}

⇒\Rightarrow

121 \over 2

=

λ2λ1{{{\lambda _2}} \over {{\lambda _1}}}

⇒\Rightarrow

λ1\lambda _1

= 2

λ2\lambda _2
Q37
A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be (h = 6.6 × \times 10 −-34 J s)
A 6.6 × \times 10 −-32 m
B 6.6 × \times 10 −-34 m
C 1.0 × \times 10 −-35 m
D 1.0 × \times 10 −-32 m
Correct Answer
Option C
Solution

According to the de-Broglie equation, λ\lambda =

hmv{h \over {mv}}

∴\therefore λ\lambda =

6.6×10−340.66×100=1×10−35 m{{6.6 \times {{10}^{ - 34}}} \over {0.66 \times 100}} = 1 \times {10^{ - 35}}\,m
Q38
Maximum number of electrons in a subshell of an atom is determined by the following
A 2ll + 1
B 4ll −- 2
C 2n 2
D 4ll + 2
Correct Answer
Option D
Solution

Maximum number of electrons in a subshell = 2(2l+1) = 4l + 2

Q39
Which of the following is not permissible arrangement of electrons in an atom ?
A n = 5, ll = 3, m = 0, s = +1/2
B n = 3, ll = 2, m = −- 3, s = −-1/2
C n = 3, ll = 2, m = −- 2, s = −-1/2
D n = 4, ll = 0, m = 0, s = −-1/2
Correct Answer
Option B
Solution

In an atom for any value of n, the values of l = 0 to (n-1) For a given value of l, the values of m l = -

ll

to 0 to +

ll

and the value of s = +1/2 or -1/2 In option(b), l = 2 and m l = -3 This is not possible. as for l = 2 min value of m l = -2

Q40
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 × \times 10 −-18 g cm s −-1 . The uncertainty in electron velocity is (mass of an electron is 9 × \times 10 −-28 g)
A 1 × \times 10 5 cm s −-1
B 1 × \times 10 11 cm s −-1
C 1 × \times 10 9 cm s −-1
D 1 × \times 10 6 cm s −-1
Correct Answer
Option C
Solution
Δv\Delta v

=

1×10−189×10−28=1.1×109cm s−1{{1 \times {{10}^{ - 18}}} \over {9 \times {{10}^{ - 28}}}} = 1.1 \times {10^9}cm\,{s^{ - 1}}
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