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Atoms and Nuclei

NEET Physics · 119 questions · Page 2 of 12 · Click an option or "Show Solution" to reveal answer

Q11
The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to (116)th \left(\dfrac{1}{16}\right)^{\text{th }} of its initial value?
A 40 minutes
B 60 minutes
C 80 minutes
D 20 minutes
Correct Answer
Option C
Solution

Half life

T=20 minT=20 \mathrm{~min}

Left fraction of activity

116\frac{1}{16}
∵RR0=(12)t/T116=(12)t/20(12)4=(12)t/204=t20t=80 min\begin{aligned} \because \frac{\mathrm{R}}{\mathrm{R}_{0}} & =\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}} \\ \frac{1}{16} & =\left(\frac{1}{2}\right)^{\mathrm{t} / 20} \\ \left(\frac{1}{2}\right)^{4} & =\left(\frac{1}{2}\right)^{\mathrm{t} / 20} \\ 4 & =\frac{\mathrm{t}}{20} \\ \mathrm{t} & =80 \mathrm{~min} \end{aligned}
Q12
In hydrogen spectrum, the shortest wavelength in the Balmer series is λ\lambda. The shortest wavelength in the Bracket series is :
A 4λ4 \lambda
B 9λ9 \lambda
C 16λ16 \lambda
D 2λ2 \lambda
Correct Answer
Option A
Solution

Shortest wavelength in Balmer series when transition of

e−e^{-}

from ∞\infty to

n=2\mathrm{n}=2
∵1λ=Rz2[122−1∞2]\because \frac{1}{\lambda}=\mathrm{Rz}^{2}\left[\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right]
1λ=R4\frac{1}{\lambda}=\frac{R}{4}

.... (1) Shortest wavelength is Bracket series when transition of

e−e^{-}

from ∞\infty to

n=4\mathrm{n}=4
1λ′=R(1)2[142−1∞2]⇒1λ′=R16\frac{1}{\lambda^{\prime}}=\mathrm{R}(1)^{2}\left[\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right] \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{\mathrm{R}}{16}

..... (2) Eq. (1) / Eq. (2)

λ′λ=R4×16R⇒λ′=4λ\frac{\lambda^{\prime}}{\lambda}=\frac{\mathrm{R}}{4} \times \frac{16}{\mathrm{R}} \Rightarrow \lambda^{\prime}=4 \lambda
Q13
The radius of inner most orbit of hydrogen atom is 5.3×10−11 m5.3 \times 10^{-11} \mathrm{~m}. What is the radius of third allowed orbit of hydrogen atom?
A 1.06 Ao\mathop A\limits^o
B 1.59 Ao\mathop A\limits^o
C 4.77 Ao\mathop A\limits^o
D 0.53 Ao\mathop A\limits^o
Correct Answer
Option C
Solution

The formula to calculate the radius of the nth orbit in a hydrogen atom is given by the equation:

rn=r0×n2r_n = r_0 \times n^2

where,

rnr_n

is the radius of the nth orbit,

r0r_0

is the radius of the innermost orbit (known as the Bohr radius), and

nn

is the orbit number. Given that the radius of the innermost orbit (

r0r_0

) is

5.3×10−11 m5.3 \times 10^{-11} \mathrm{~m}

or equivalently

0.53Ao0.53 \mathop A\limits^o

, and we're interested in finding the radius of the third orbit (

n=3n = 3

). So,

r3=0.53×(3)2r_3 = 0.53 \times (3)^2
Ao\mathop A\limits^o
r3=0.53×9r_3 = 0.53 \times 9
Ao\mathop A\limits^o
r3=4.77r_3 = 4.77
Ao\mathop A\limits^o

Therefore, the radius of the third allowed orbit of the hydrogen atom is 4.77

Ao\mathop A\limits^o

.

Q14
Let R 1 be the radius of the second stationary orbit and R 2 be the radius of the fourth stationary orbit of an electron in Bohr's model. The ratio R1R2{{{R_1}} \over {{R_2}}} is :
A 4
B 0.25
C 0.5
D 2
Correct Answer
Option B
Solution

Radius of Bohr's orbit depends on principal quantum number (n) as

R∝n2R \propto {n^2}

Now,

R1R2=(2)2(4)2=14=0.25{{{R_1}} \over {{R_2}}} = {{{{(2)}^2}} \over {{{(4)}^2}}} = {1 \over 4} = 0.25
Q15
At any instant, two elements X 1 and X 2 have same number of radioactive atoms. If the decay constant of X 1 and X 2 are 10 λ\lambda and λ\lambda respectively, then the time when the ratio of their atoms becomes 1e{1 \over e} respectively will be :
A 15λ{1 \over {5\lambda }}
B 111λ{1 \over {11\lambda }}
C 19λ{1 \over {9\lambda }}
D 16λ{1 \over {6\lambda }}
Correct Answer
Option C
Solution

Number of radioactive nuclei at any time is

N=N0e−λtN = {N_0}{e^{ - \lambda t}}

Initial number of nuclei is same for sample X 1 & X 2 . Let after time 't' the ratio of their atoms become

1e{1 \over e}
⇒N0e−10λ×tN0e−λ×t=1e⇒e−9λt=e−1⇒t=19λ\Rightarrow {{{N_0}{e^{ - 10\lambda \times t}}} \over {{N_0}{e^{ - \lambda \times t}}}} = {1 \over e} \Rightarrow {e^{ - 9\lambda t}} = {e^{ - 1}} \Rightarrow t = {1 \over {9\lambda }}

Hence,

t=19λt = {1 \over {9\lambda }}
Q16
The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is 2.4 ×\times 10 39 . The ratio of the proportionality constant, K=14πε0K = {1 \over {4\pi {\varepsilon _0}}} to the gravitational constant G is nearly (Given that the charge of the proton and electron each = 1.6 ×\times 10 −-19 C, the mass of the electron = 9.11 ×\times 10 −-31 kg, the mass of the proton = 1.67 ×\times 10 −-27 kg) :
A 10
B 10 20
C 10 30
D 10 40
Correct Answer
Option B
Solution

Ratio of magnitude of Coulomb's electrostatic force to the gravitational force

FEFG=(Kq1q2r2)(Gm1m2r2)=Kq1q2Gm1m2{{{F_E}} \over {{F_G}}} = {{\left( {{{K{q_1}{q_2}} \over {{r^2}}}} \right)} \over {\left( {{{G{m_1}{m_2}} \over {{r^2}}}} \right)}} = {{K{q_1}{q_2}} \over {G{m_1}{m_2}}}
⇒2.4×1039=KG×1.6×10−19×1.6×10−199.11×10−31×1.67×10−27\Rightarrow 2.4 \times {10^{39}} = {K \over G} \times {{1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}} \over {9.11 \times {{10}^{ - 31}} \times 1.67 \times {{10}^{ - 27}}}}
⇒2.4×1039=KG×2.5615.21×1020⇒KG=14.26×1019\Rightarrow 2.4 \times {10^{39}} = {K \over G} \times {{2.56} \over {15.21}} \times {10^{20}} \Rightarrow {K \over G} = 14.26 \times {10^{19}}
⇒KG=1.426×1020\Rightarrow {K \over G} = 1.426 \times {10^{20}}

⇒\Rightarrow Ratio ≈\approx 10 20

Q17
In the given nuclear reaction, the element X is 1122Na→X+e++v{}_{11}^{22}Na \to X + {e^ + } + v
A 1123Na{}_{11}^{23}Na
B 1023Ne{}_{10}^{23}Ne
C 1022Ne{}_{10}^{22}Ne
D 1222Mg{}_{12}^{22}Mg
Correct Answer
Option C
Solution

The nuclear reaction is given as

1122Na→ZAX++1e0+v{}_{11}^{22}Na \to {}_Z^AX + {}_{ + 1}{e^0} + v

From conservation of atomic number

11=Z+1⇒Z=10⇒Ne11 = Z + 1 \Rightarrow Z = 10 \Rightarrow Ne

From conservation of mass number

22=A+0⇒A=2222 = A + 0 \Rightarrow A = 22

∴\therefore

ZAX=1022Ne{}_Z^AX = {}_{10}^{22}Ne
Q18
Let T 1 and T 2 be the energy of an electron in the first and second excited states of hydrogen atoms, respectively. According to the Bohr's model of an atom, the ratio T 1 : T 2 is
A 1 : 4
B 4 : 1
C 4 : 9
D 9 : 4
Correct Answer
Option D
Solution
En=E0n2{E_n} = {{{E_0}} \over {{n^2}}}

, For first excited state ⇒\Rightarrow n = 2 For second excited state ⇒\Rightarrow n = 3

⇒T1T2=E04E09=94\Rightarrow {{{T_1}} \over {{T_2}}} = {{{{{E_0}} \over 4}} \over {{{{E_0}} \over 9}}} = {9 \over 4}
Q19
A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is
A 1 : 1
B 4 : 5
C 5 : 4
D 25 : 16
Correct Answer
Option C
Solution

Radius of nuclei with mass number A varies as

R=R0A1/3R = {R_0}{A^{1/3}}
R1R2=(12564)1/3=54=5:4{{{R_1}} \over {{R_2}}} = {\left( {{{125} \over {64}}} \right)^{1/3}} = {5 \over 4} = 5:4
Q20
A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding per nucleon of unfragmented nuclei is 7.6MeV while that of fragments is 8.5MeV. The total gain in the Binding Energy in the process is :
A 216MeV
B 0.9MeV
C 9.4MeV
D 804MeV
Correct Answer
Option A
Solution

Given binding energy per nucleon of X, Y & Z are 7.6MeV, 8.5MeV & 8.5MeV respectively, Gain in binding energy is : Q = Binding Energy of products −- Binding energy of reactants = (120 ×\times 8.5 ×\times 2) −- (240 ×\times 7.6) MeV = 216 MeV

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