Book Chosen

Physics Part II

Subject Chosen

Physics

Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12

Assume that a neutron breaks into a proton and an electron. The energy released during this process is(Mass of neutron = 1.6725 x 10–27kg; mass of proton = 1.6725 x 10–27kg; mass of electron = 9 x 10–31kg)

  • 0.73 MeV

  • 7.10 MeV

  • 6.30 MeV

  • 5.4 MeV


A.

0.73 MeV

increment straight m space equals space left parenthesis straight m subscript straight p plus straight m subscript straight e right parenthesis minus straight m subscript straight n
space equals space 9 space straight x 10 to the power of negative 31 end exponent space kg.
Energy space released space equals space left parenthesis 9 space straight x space 10 to the power of negative 31 end exponent space kg right parenthesis space straight c squared space joules
space equals space fraction numerator 9 space straight x 10 to the power of negative 31 end exponent space straight x space left parenthesis 3 space straight x space 10 to the power of 8 right parenthesis squared over denominator 1.6 space straight x space 110 to the power of negative 13 end exponent end fraction space space MeV
space equals space 0.73 space MeV
increment straight m space equals space left parenthesis straight m subscript straight p plus straight m subscript straight e right parenthesis minus straight m subscript straight n
space equals space 9 space straight x 10 to the power of negative 31 end exponent space kg.
Energy space released space equals space left parenthesis 9 space straight x space 10 to the power of negative 31 end exponent space kg right parenthesis space straight c squared space joules
space equals space fraction numerator 9 space straight x 10 to the power of negative 31 end exponent space straight x space left parenthesis 3 space straight x space 10 to the power of 8 right parenthesis squared over denominator 1.6 space straight x space 110 to the power of negative 13 end exponent end fraction space space MeV
space equals space 0.73 space MeV
1351 Views

(a) Draw the plot of binding energy per nucleon (BE/A) as a function of mass number A. Write two important conclusions that can be drawn regarding the nature of nuclear force.

(b) Use this graph to explain the release of energy in both the processes of nuclear fusion and fission.

(c) Write the basic nuclear process of neutron undergoing –decay. Why is the detection of neutrinos found very difficult?


Graphical representation of (BE/A) for nucleons with mass number A.

The variation of binding energy per nucleon VS. mass number is shown in the figure:

Characteristics of Nuclear force:

(i) Nuclear forces non-central and short ranged force.

(ii) Nuclear forces between proton-neutron and neutron-neutron are strong and attractive in nature.

b) When a heavy nucleus (A > 235 say) breaks into two lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e, nucleons get more tightly bound. This implies that energy would be released in nuclear fission.

When two very light nuclei (A £10) join to form a heavy nucleus, the binding is energy per nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion.

c) During the decay process of neutron, we have

n presubscript 0 presuperscript 1 space rightwards arrow space p presubscript 1 presuperscript 1 space plus space beta presubscript negative 1 end presubscript presuperscript 0 space plus space nu with bar on top space 

Neutrinos show weak interaction with other particles. Hence, its detection is very different.

2822 Views

Proton, Deuteron and alpha particle of the same kinetic energy is moving in circular trajectories in a constant magnetic field. The radii of the proton, deuteron and alpha particle are respectively rp, rd and rα. Which one of the following relations is correct?

  •  rα = rp= rd

  •  rα = rp< rd

  •  rα > rd> rp

  •  rα = rd> rp


B.

 rα = rp< rd

For charged particle moving with a speed v, in magnetic field B, on a circular track of radius

straight r space equals space mv over qB space equals space fraction numerator square root of 2 km end root over denominator qB end fraction
open parentheses therefore space mv space equals space straight p space and space straight K space equals fraction numerator straight p squared over denominator 2 straight m end fraction space rightwards double arrow space straight p space equals space square root of 2 km end root close parentheses
rightwards double arrow space straight r space proportional to space fraction numerator square root of straight m over denominator straight q end fraction
or space straight m subscript straight d space equals 2 straight m subscript straight p space and space straight q subscript straight d space equals space straight q subscript straight p semicolon
straight m subscript straight alpha space equals space 4 straight m subscript straight p space and space straight q subscript straight alpha space equals space 2 straight q subscript straight p
rightwards double arrow straight r subscript straight p colon straight r subscript straight d colon straight r subscript straight alpha space equals space fraction numerator square root of straight m subscript straight p end root over denominator straight q subscript straight p end fraction colon fraction numerator square root of 2 straight m subscript straight p end root over denominator straight q subscript straight p end fraction colon fraction numerator square root of 4 straight m subscript straight p end root over denominator 2 straight q subscript straight p end fraction
space equals space 1 colon space square root of 2 colon 1
straight r subscript straight alpha space equals space straight r subscript straight p space less than thin space straight r subscript straight d

For charged particle moving with a speed v, in magnetic field B, on a circular track of radius

straight r space equals space mv over qB space equals space fraction numerator square root of 2 km end root over denominator qB end fraction
open parentheses therefore space mv space equals space straight p space and space straight K space equals fraction numerator straight p squared over denominator 2 straight m end fraction space rightwards double arrow space straight p space equals space square root of 2 km end root close parentheses
rightwards double arrow space straight r space proportional to space fraction numerator square root of straight m over denominator straight q end fraction
or space straight m subscript straight d space equals 2 straight m subscript straight p space and space straight q subscript straight d space equals space straight q subscript straight p semicolon
straight m subscript straight alpha space equals space 4 straight m subscript straight p space and space straight q subscript straight alpha space equals space 2 straight q subscript straight p
rightwards double arrow straight r subscript straight p colon straight r subscript straight d colon straight r subscript straight alpha space equals space fraction numerator square root of straight m subscript straight p end root over denominator straight q subscript straight p end fraction colon fraction numerator square root of 2 straight m subscript straight p end root over denominator straight q subscript straight p end fraction colon fraction numerator square root of 4 straight m subscript straight p end root over denominator 2 straight q subscript straight p end fraction
space equals space 1 colon space square root of 2 colon 1
straight r subscript straight alpha space equals space straight r subscript straight p space less than thin space straight r subscript straight d

808 Views

Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be:

  • 1: 16

  • 4 : 1

  • 1: 4

  • 5: 4


D.

5: 4

Given 80 min = 4 half-lives of A = 2 half-lives of B.
Let the initial number of nuclei in each sample be N.

For radioactive element A,

NA after 80 min = N/24

⇒ Number of A nuclides decayed  =straight N minus straight N over 16 space equals space 15 over 16 straight N


For radioactive element B,

NB after 80 min  = N/22
⇒ Number of B nuclides decayed
straight N minus straight N over 4 space equals space 3 over 4 straight N
therefore, the ratio of decayed numbers of A and B nuclei will be

fraction numerator open parentheses 15 divided by 16 close parentheses straight N over denominator left parenthesis 3 divided by 4 right parenthesis straight N end fraction space equals space 5 over 4

Given 80 min = 4 half-lives of A = 2 half-lives of B.
Let the initial number of nuclei in each sample be N.

For radioactive element A,

NA after 80 min = N/24

⇒ Number of A nuclides decayed  =straight N minus straight N over 16 space equals space 15 over 16 straight N


For radioactive element B,

NB after 80 min  = N/22
⇒ Number of B nuclides decayed
straight N minus straight N over 4 space equals space 3 over 4 straight N
therefore, the ratio of decayed numbers of A and B nuclei will be

fraction numerator open parentheses 15 divided by 16 close parentheses straight N over denominator left parenthesis 3 divided by 4 right parenthesis straight N end fraction space equals space 5 over 4
385 Views

The half life of a radioactive substance is 20 minutes. The approximate time interval (t2 - t1) between the time t2 when 2/3 of it has decayed and time t1 when 1/3 of it had decayed is

  • 14 min

  • 20 min

  • 28 min

  • 7 min 


B.

20 min

2 over 3 straight N subscript 0 space equals space straight N subscript 0 straight e to the power of negative λt subscript 1 end exponent
1 third straight N subscript 0 space equals space straight N subscript 0 straight e to the power of negative λt subscript 2 end exponent
2 space equals straight e to the power of straight lambda left parenthesis space straight t subscript 2 minus straight t subscript 1 right parenthesis end exponent
straight lambda left parenthesis straight t subscript 2 minus straight t subscript 1 right parenthesis space equals space l n 2
left parenthesis straight t subscript 2 minus straight t subscript 1 right parenthesis space equals space fraction numerator l straight n begin display style 2 end style over denominator straight lambda end fraction space equals space 20 space min
2 over 3 straight N subscript 0 space equals space straight N subscript 0 straight e to the power of negative λt subscript 1 end exponent
1 third straight N subscript 0 space equals space straight N subscript 0 straight e to the power of negative λt subscript 2 end exponent
2 space equals straight e to the power of straight lambda left parenthesis space straight t subscript 2 minus straight t subscript 1 right parenthesis end exponent
straight lambda left parenthesis straight t subscript 2 minus straight t subscript 1 right parenthesis space equals space l n 2
left parenthesis straight t subscript 2 minus straight t subscript 1 right parenthesis space equals space fraction numerator l straight n begin display style 2 end style over denominator straight lambda end fraction space equals space 20 space min
580 Views