Practice Start Test

Atoms and Nuclei

NEET Physics · 119 questions · Page 4 of 12 · Click an option or "Show Solution" to reveal answer

Q31
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
A 1
B 4
C 0.5
D 2
Correct Answer
Option B
Solution

In case of Balmer series : n 1 = ∞\infty, n 2 = 2 ∴\therefore

1λB=Rc(122−1∞2){1 \over {{\lambda _B}}} = Rc\left( {{1 \over {{2^2}}} - {1 \over {{\infty ^2}}}} \right)

=

Rc4{{Rc} \over 4}

In case of Lyman series : n 1 = ∞\infty, n 2 = 1 ∴\therefore

1λL=Rc(112−1∞2){1 \over {{\lambda _L}}} = Rc\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)

= Rc ∴\therefore

λBλL{{{\lambda _B}} \over {{\lambda _L}}}

=

41{4 \over 1}

= 4

Q32
The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is
A 15
B 30
C 45
D 60
Correct Answer
Option D
Solution

N 0 = Nuclei at time t = 0 N 1 = Remaining nuclei after 40% decay = (1 – 0.4) N 0 = 0.6 N 0 N 2 = Remaining nuclei after 85% decay = (1 – 0.85) N 0 = 0.15 N 0 ∴\therefore

N2N1{{{N_2}} \over {{N_1}}}

=

0.15N00.6N0{{0.15{N_0}} \over {0.6{N_0}}}

=

(12)2{\left( {{1 \over 2}} \right)^2}

Hence number of half-lives is 2, so the time needed is : t = 2 × 30 = 60 mints

Q33
If an electron in a hydrogen atom jumps from the 3 rd orbit to the 2 nd orbit, it emits a photon of wavelength λ\lambda . When it jumps from the 4 th orbit to the 3 rd orbit, the corresponding wavelength of the photon will be
A 1625λ{{16} \over {25}}\lambda
B 916λ{9 \over {16}}\lambda
C 207λ{{20} \over 7}\lambda
D 2013λ{{20} \over {13}}\lambda
Correct Answer
Option C
Solution

When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,

1λ=R(1nf2−1ni2){1 \over \lambda } = R\left( {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right)

On jumping from 3 rd orbit to 2 nd orbit,

1λ=R(122−132){1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)

=

5R36{{5R} \over {36}}

On jumping from 4 th orbit to 3 rd orbit,

1λ′=R(132−142){1 \over {\lambda '}} = R\left( {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right)

=

7R144{{7R} \over {144}}

∴\therefore λ\lambda' =

1447×5λ36{{144} \over 7} \times {{5\lambda } \over {36}}

=

20λ7{{20\lambda } \over 7}
Q34
When an α\alpha -particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as
A 1m2{1 \over {{m^2}}}
B m
C 1m{1 \over m}
D 1m{1 \over {\sqrt m }}
Correct Answer
Option C
Solution

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.

Kinetic energy =

12mv2{1 \over 2}m{v^2}

Potential energy =

KQqr{{KQq} \over r}

∴\therefore

12mv2{1 \over 2}m{v^2}

=

KQqr{{KQq} \over r}

⇒\Rightarrow r ∝\propto

1m{1 \over m}
Q35
Given the value of Rydberg constant is 10 7 m −-1 , the wave number of the last line of the Balmer series in hydrogen spectrum will be
A 0.25 × \times 10 7 m −-1
B 2.5 × \times 10 7 m −-1
C 0.025 × \times 10 4 m −-1
D 0.5 × \times 10 7 m −-1
Correct Answer
Option A
Solution

The wave number of the last line of the Balmer series in hydrogen spectrum is given by

1λ=R(122−1∞2){1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{\infty ^2}}}} \right)

=

R4{R \over 4}

=

1074{{{{10}^7}} \over 4}

= 0.25 ×\times 10 7 m −-1

Q36
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then
A The helium nucleus has more momentum than the thorium nucleus.
B The helium nucleus has less kinetic energy than the thorium nucleus.
C The helium nucleus has more kinetic energy than the thorium nucleus.
D The helium nucleus has less momentum than the thorium nucleus.
Correct Answer
Option C
Solution

In an explosion a body breaks up into two pieces of unequal masses both part will have numerically equal momentum and lighter part will have more velocity.

U →\to Th + He KE Th =

P22mTh{{{P^2}} \over {2{m_{Th}}}}

KE He =

P22mHe{{{P^2}} \over {2{m_{He}}}}

since m He is less so KE He will be more.

Q37
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
A 275{{27} \over 5}
B 527{5 \over {27}}
C 49{4 \over 9}
D 94{9 \over 4}
Correct Answer
Option B
Solution

For the longest wavelength in the Lyman series, n 1 = 1 and n 2 = 2

1λL=R(112−122){1 \over {{\lambda _L}}} = R\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)

=

3R4{{3R} \over 4}

For the longest wavelength in the Balmer series, n 1 = 2 and n 2 = 3

1λB=R(122−132){1 \over {{\lambda _B}}} = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)

=

5R36{{5R} \over {36}}

∴\therefore

λLλB=43R365R{{{\lambda _L}} \over {{\lambda _B}}} = {{{4 \over {3R}}} \over {{{36} \over {5R}}}}

=

527{5 \over {27}}
Q38
Consider 3 rd orbit of He + (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 × \times 10 9 constant, Z = 2 and h (Planck's Constant) = 6.6 × \times 10 −-34 J s]
A 0.73 × \times 10 6 m/s
B 3.0 × \times 10 8 m/s
C 2.92 × \times 10 6 m/s
D 1.46 × \times 10 6 m/s
Correct Answer
Option D
Solution

Energy of electron in He + 3 rd orbit E 3 = -13.6 ×\times

49eV{4 \over 9}eV

= -13.6 ×\times

49{4 \over 9}

×\times 1.6 ×\times 10 -19 J = - 9.7 ×\times 10 -19 J According to Bohr’s model, Kinetic energy of electron in the 3 rd orbit = – E 3 ∴\therefore 9.7 ×\times 10 -19 =

12mev2{1 \over 2}{m_e}{v^2}

v =

2×9.7×10199.1×10−31\sqrt {{{2 \times 9.7 \times {{10}^{19}}} \over {9.1 \times {{10}^{ - 31}}}}}

= 1.46 ×\times 10 6 m/s

Q39
If radius of the 1327{}_{13}^{27} Al nucleus is taken to be R Al , then the radius of 53125{}_{53}^{125}Te nucleus is nearly
A 35RAl{3 \over 5}{R_{Al}}
B (1353)1/3RAl{\left( {{{13} \over {53}}} \right)^{1/3}}{R_{Al}}
C (5313)1/3RAl{\left( {{{53} \over {13}}} \right)^{1/3}}{R_{Al}}
D 53RAl{5 \over 3}{R_{Al}}
Correct Answer
Option D
Solution

As we know, R = R 0 (A) 1/3 where A = mass number R Al = R 0 (27) 1/3 = 3R 0 R Te = R 0 (125) 1/3 = 5R 0 =

53RAl{5 \over 3}{R_{Al}}
Q40
Hydrogen atom in ground state is excited by a monochromatic radiation of λ\lambda = 975 A∘\mathop A\limits^ \circ . Number of spectral lines in the resulting spectrum emitted will be
A 3
B 2
C 6
D 0 10
Correct Answer
Option C
Solution

Energy of radiation : E =

12375975{{12375} \over {975}}

= 12.7 eV After absorbing a photon of energy 12.75 eV, the electron will reach to third excited state of energy –0.85 eV, since energy difference corresponding to n = 1 and n = 4 is 12.75 eV.

∴\therefore Number of spectral lines emitted =

n(n−1)2=4(4−1)2{{n\left( {n - 1} \right)} \over 2} = {{4\left( {4 - 1} \right)} \over 2}

= 6

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