For a radioactive material , half-life is 10 minutes . If initially there are 600 number of nuclei…

Q: For a radioactive material , half-life is 10 minutes . If initially there are 600 number of nuclei , the time taken (in minutes) for the disintegration of 450 nuclei is

(a) 30

(b) 15

(c) 10

(d) 20

Ans: (d)

Sol: N = 600 – 450 = 150

N/N0 = (1/2)n

150/600 = (1/2)n

1/4 = (1/2)n

n = 2

Number of half life = 2

t = n T1/2

t = 2 × 10

The order of energies of energy levels, A, B and C is EA < EB < EC , if the wavelength corresponding to transitions...

Q: The order of energies of energy levels, A, B and C is EA < EB < EC , if the wavelength corresponding to transitions C → B, B → A and C → A are λ1 , λ2 and λ3 respectively , then which of the following relations is correct ?

(a) λ1 + λ2 + λ3 = 0

(b) λ32 = λ12 + λ22

(c) λ3 = λ1 + λ2

(d) $\displaystyle \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} $

Ans: (d)

Sol: ECA = ECB + EBA

$\displaystyle \frac{h c}{\lambda_3} = \frac{h c}{\lambda_1} + \frac{h c}{\lambda_2} $

$\displaystyle \frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} $

$ \displaystyle \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} $

A deutron of atomic mass 2·0147 a.m.u. and negligible kinetic energy is absorbed by a Li6 nucleus…

Q: A deutron of atomic mass 2·0147 a.m.u. and negligible kinetic energy is absorbed by a Li6 nucleus of mass 6·0169 a.m.u., the intermediate nucleus disintegrates spontaneously into two α – particles, each of mass 4·0039 a.m.u. The energy transferred to each α particle is

(a) 12·08 MeV

(b) 11.08 MeV

(c) 6·04 MeV

(d) 5·54 MeV

Ans: (b)

Sol: The given nuclear reaction is

3Li6 + 1H2  → 2 (2He4 )

Mass defect, Δm = (6.0169+2.0147)-2×4.0039

=(8.0316-8.0078) amu

Δm = 0.0238 amu

Energy available = 0.0238 × 931 MeV

KE of each α particle =(0.0238×931)/2 MeV

= 11.08 MeV

A radioactive source, in the form of a metallic sphere of radius 10−2 m emits β particle at the rate of…

Q:A radioactive source, in the form of a metallic sphere of radius 10−2 m emits β particle at the rate of 1011 particle per second. The source is electrically insulated. The time required for its potential to be raised by 4 V assuming 60 % of emitted β particles escape the source is

(a) 7·29 × 10−4 s

(b) 2·79 × 10−4 s

(c) 4.63 × 10−4 s

(d) 5·67 × 10−4 s

Ans: (c)

Sol: Here, r = 10-2 m , t = ? V = 4 volt.

Number of β particle escaping the sphere/sec

$\displaystyle n = \frac{60}{100}\times 10^{11} $

n = 6 × 1010 C/sec

Charge on each β particle, e= 1.6 × 10-19 C

Charge acquired by the sphere in one second

q = ne = 6 × 1010×1.6 × 10-19

= 9.6 × 10-19 C

Total charge acquired Q = CV = 4 π ∈0 r (V)

$ \displaystyle = \frac{1}{9\times 10^9} \times 10^{-2}\times 4 $

$ \displaystyle Q = \frac{4}{9} \times 10^{-11} C $

Time required $ \displaystyle = \frac{Q}{q} $

$ \displaystyle = \frac{4}{9}\times 10^{-11}/9.6\times 10^{-9} $

= 4.63×10-4 s

A nucleus with mass number 220 initially at rest, emits an alpha particle. If the Q value of the reaction…

Q: A nucleus with mass number 220 initially at rest, emits an alpha particle. If the Q value of the reaction is 5·5 MeV, calculate the K.E. of the alpha particles.

(a) 4.4 MeV

(b) 5.4 MeV

(c) 5·6 MeV

(d) 6·5 MeV

Ans:(b)

Sol: From conservation of linear momentum,

pi = pf

As Linear Momentum $ \displaystyle p = \sqrt{2 m E}$

$ \displaystyle 0 = \sqrt{2 (216 m) E_1} – \sqrt{2 (4 m) E_2} $ (since , moves in opposite direction )

$ \displaystyle \sqrt{2 (216 m) E_1} = \sqrt{2 (4 m) E_2} $

E2 = 54 E1

As, E1 + E2 = 5.5 MeV

E1 + 54 E1 = 5.5 MeV

55 E1 = 5.5 MeV

E1 = 0.1 MeV

Hence , E2 = 5.4 MeV