Question Set C

CBSE 12th Standard Physics Dual Nature of Radiation and Matter HOT Questions Study Material - QB365

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Questions:

Multiple Choice Question

8 x 1 = 8

The maximum K.E of photoelectrons ejected from a photometer when it is irradiated with radiation of wavelength 400nm is 1eV. If the threshold energy of the surface is 1.9eV.

(a)

the maximum K.E. of photoelectrons when it is irradiated with 500nm photons will be 0.42eV

(b)

the maximum K.E. of photoelectrons when it is viradiated with 500nm photons will be 0.42eV

(c)

maximum K.E will increase if the intensity of radiation is increased

(d)

the longest wavelength which will eject the photoelectron from the surface is nearly 610nm

Nuclear reactor in which uranium -235 is used as fuel, uses 2 kg of uranium -235 in 30 days. Then power output of the reactor will be (given : Energy released per fission = 185 MeV)

(a)

43.5 MW

(b)

58.5 MW

(c)

69.6 MW

(d)

73.1 MW

The kinetic energy of an electron is E.When the incident light has wavelength \(\lambda \) To increase the kinetic energy to 2E, the incident light must have wavelength.

(a)

\(\frac { hc }{ E\lambda -hc } \)

(b)

\(\frac { h\lambda }{ E\lambda +hc } \)

(c)

\(\frac { hc\lambda }{ E\lambda +hc } \)

(d)

\(\frac { hc\lambda }{ E\lambda -hc } \)

Radioactive radiations in order of increasing penetrating power are:

(a)

\(\gamma ,\beta ,\alpha \quad \)

(b)

\(\gamma ,\alpha ,\beta \)

(c)

\(\alpha ,\beta ,\gamma \)

(d)

\(\alpha ,\gamma ,\beta \)

The slope of frequency of incident light and stopping potential for a given surface will be

(a)

(b)

h/e

(c)

(d)

The simple Bohr modle is not applicable to \({ He }^{ 4 }\) atom because

(a)

\({ He }^{ 4 }\) is an inert gas

(b)

\({ He }^{ 4 }\) has neutrons in the nucleus

(c)

\({ He }^{ 4 }\) has one more electron

(d)

electrons are not subject to central forces

Relativistic corrections become necessary when the expression for the kinetic energy \(\frac { 1 }{ 2 } m{ v }^{ 2 }\) becomes comparable with \(m{ c }^{ 2 }\) where m is the mass of the particle. At what de-Broglie wavelength will relativistic corrections become important for an electron?

(a)

\(\lambda =10nm\)

(b)

\(\lambda ={ 10 }^{ -1 }nm\)

(c)

\(\lambda ={ 10 }^{ -4 }nm\)

(d)

\(\lambda ={ 10 }^{ -6 }nm\)

An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state

(a)

the electron would not move in circular orbits

(b)

the energy would be (2)⁴ times that of a H-atom

(c)

the electrons, orbit would go around the protons

(d)

the molecule will soon decay in a proton and a H-atom.

Generic 2

3 x 2 = 6

The equivalent wavelength of a moving electron has the same value as that of a photon having an energy of 6 x 10^-17 J. Calculate the momentum of the electron.

10.

Calculate
(i) momentum and
(ii) de-Broglie wavelength of the electron accelerated through a potential difference of 56 V

11.

The natural boron is found to be composed of two isotopes of \(_{ 5 }{ B^{ 10 } }\) and \(_{ 5 }{ B^{ 11 } }\) . The masses of these two isotopes are 10.003 u and 11.009 u respectively. The atomic mass of natural boron is 10.81 u. Determine the relative abundance of each isotope in the natural boron.

Generic 3

2 x 3 = 6

12.

The neutron separation energy is defined to be the energy required to remove a neutron from a nucleus. Obtain the neutron separation energy of the nuclei \(_{ 20 }{ Ca^{ 41 } }and \ _{ 13 }{ Al^{ 27 } }\) from the following data:
\(m(_{ 20 }{ Ca^{ 40 }) }=39.962591u\ and\ m(_{ 20 }{ Ca^{ 41 } })=40.962278\)
\(m(_{ 13 }{ Al^{ 26 }) }=25.986895u\ and\ m(_{ 13 }{ Al^{ 27 } })=26.981541u\)

13.

M1 and M2 represent the masses of \(_{ 10 }{ { Ne }^{ 20 } }\)nucleus and \(_{ 20 }{ { Ca }^{ 40 } }\) nucleus respectively. State whether M2=2M1 or M2>2M1 or M2<2M1

Generic 5

4 x 5 = 20

14.

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.
Use this graph to explain the release of energy in both the processes of nuclear fusion and fission.
Write the basic nuclear process of neutron undergoing p-decay. Why is the detection of neutrinos found very difficult?

15.

Ram knows that red light has greater and so it is much bright, but in case of photoelectric emission it cannot produce the emission of electrons from a clean zinc surface, while even weak ultraviolet radiation can do so. He could not know specific cause of such thing. Then he went to his friend Shyam for its specific explanation. Shyam explained him that the photoemission of electron does not depend on the intensity while it depends on the frequency and thus on the energy of photon of incident light. The energy of photon of red light cannot emit photoelectrons. Similarly, the energy of photon of ultraviolet light is greater than the work function of zinc, so ultraviolet light can emit photoelectrons.
(a) What values are noticed in Shyam?
(b) The work functions of lithium and copper are 2.3eV and 4eV respectively. Which of these metals are useful for the photoelectric cell working with visible light? Explain.

16.

Find DE-Broglie wavelength of neutron at\(127°C\). Given Boltzmann constant, \(K=1.38\times { 10 }^{ -23 }J{ mole }^{ -1 }{ K }^{ -1 },h=6.63\times { 10 }^{ -34 }Js,\)mass of neutron = \(1.66\times { 10 }^{ -27 }kg\)

17.

(a) The work function for the surface of aluminum is 4.2 eV. How much potential difference will be required to stop the emission of maximum energy electrons emitted by light of \(2000\overset { \circ }{ A } \) wavelength?
(b) What will be the wavelength of that incident light for which stopping potential will be zero? \(h=6.63\times { 10 }^{ -34 }Js,c=3\times { 10 }^{ 8 }{ ms }^{ -1 }\)

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Answers Key:

Multiple Choice Question

8 x 1 = 8

(b)
58.5 MW

(b)
h/e

(d)
\(\lambda ={ 10 }^{ -6 }nm\)

Generic 2

3 x 2 = 6

E = Energy of the photon = hv = \(\frac { hc }{ \lambda } \)
\(\lambda =\frac { hc }{ E } \)
Wave length of the moving electron \(=\lambda =\frac { hc }{ E } 1/2\)
Movement of the electron = p
\(=\frac { h }{ \lambda } =\frac { hE }{ hc } =\frac { E }{ c } \)
\(=\frac { 6\times { 10 }^{ -17 } }{ 3\times { 10 }^{ 8 } } { kgms }^{ -1 }=2\times { 10 }^{ -25 }{ kgms }^{ -1 }\)

10.

Protential difference V = 56 V
(i)Use the kinetic energy
eV = 1/2mv²
= 2eV/m
= v²
v = \(\sqrt { \frac { 2eV }{ m } } \)
where m is mass v is velocity
p = mv = m\(\sqrt { \frac { 2eV }{ m } } \)
\(=\sqrt { 2\times 1.6\times 10^{ -19 }\times 56\times 9\times 10^{ -31 } }\)
\( =4.02\times 10^{ -24 }\ kg-m/s\)

11.

Let \(_{ 5 }{ B^{ 10 } }\) be x%. Therefore \(_{ 5 }{ B^{ 11 } }\) will be (100-x)%.
As average atomic mass = weighted average of the masses of isotopes
\(\therefore \ 10.81=\frac { 10.003x+11.009(100-x) }{ 100 } \)
Calculate x = 19.78%
\(\therefore \) (100-x)=100-19.78=80.22%

Generic 3

2 x 3 = 6

12.

When a neutron is separated from \(_{ 20 }{ Ca^{ 41 } }\), we are left with \(_{ 20 }{ Ca^{ 40 } }\)i.e., \(_{ 20 }{ Ca^{ 41 } }\rightarrow _{ 20 }{ Ca^{ 40 } }+_{ 0 }{ n^{ 1 } }\)
Now, mass defect, \(\Delta m=m(_{ 20 }{ Ca^{ 40 } })+{ m }_{ n }-m(_{ 20 }{ Ca^{ 41 } })\)
= 39.962591 + 1.008665 - 40.962278 = 0.008978 a.m.u.
Neutron separation energy = \(_{ 13 }{ Al^{ 27 } }\rightarrow _{ 13 }{ Al^{ 26 } }+_{ 0 }{ n^{ 1 } }\)
\( \Delta m=m(_{ 13 }{ Al^{ 26 } })+{ m }_{ n }-m(_{ 13 }{ Al^{ 27 } })\)
= 25.986895 + 1.008665 - 26.981541 = 0.0138454u
Neutron separation energy = 0.0138454\(\times 931\ MeV\)= 12.89MeV

13.

M2>2M1

Generic 5

4 x 5 = 20

14.

While drawing the plot. we have to keep in mind that first binding energy will increase sharply and then it will be constant almost.
For plot of binding energy per nucleon as the function of mass number A
Following are the two conclusions that can be drawn regarding the nature of the nuclear force.
The force is attractive and strong enough to produce a binding energy of few MeV per nucleon.
The two important conclusions regarding the nature of nuclear force are given below.
(i) The nuclear force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.
(ii) The constancy of the binding energy in the wide range of mass number 30 < A < 170 indicate that nuclear force is a short-range force.
(b) (i) According to the binding energy curve, a very heavy nucleus (A > 170), has lower binding energy per nucleon compared to nuclei of middle mass number (30 < A < 170).
Thus, if a heavy nucleus breaks into two nuclei of mass number between 30 and 170, nucleons get more tightly bound. This implies energy would be released in the process. (nuclear fission)
(ii) When two light nuclei (A < 10) join to form a heavier nucleus, the binding energy per nucleon of fused heavier nucleus increases.
Again it indicates that energy would be released in the process (nuclear fusion).
(c) The basic nuclear process of neutron undergoing β-decay is given as
\(n \rightarrow p+e^{-}+\bar{v}\)
Here \(\bar{v}\) is antinutrino.
Neutrino and antineutrino both are neutral particles with very small (possibly, even zero) mass compared to the electrons. They have only weak interaction with other particles. Therefore, the detection of neutrinos is found very difficult.

15.

(a) The values noticed in Shyam are:
(i) High degree of general awareness.
(ii) Concern for his friend.
(iii) Helping and caring nature.
(b) The threshold wavelength, \({ \lambda }_{ 0 }=\frac { hc }{ W } \)
For lithium, \({ \lambda }_{ 0 }=\frac { 12375 }{ 2.3 } \overset { 0 }{ A } =5380\overset { 0 }{ A } \)
For copper, \({ \lambda }_{ 0 }=\frac { 12375 }{ 4 } \overset { 0 }{ A } =3094\overset { 0 }{ A } \)
The wavelength 5380\(\overset { 0 }{ A } \) lies in visible region, thus lithium will be useful for photoelectric cell.

16.

Here,
\(T=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\)
Energy of neutron at127° C,
\(E=\frac{3}{2} k T=\frac{3}{2} \times 1.38 \times 10^{-23} \times 400\)
= 8.28 x 10^-21 J
\(\therefore \quad \lambda=\frac{h}{\sqrt{2 m E}}\)
\(=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.66 \times 10^{-27} \times 8.28 \times 10^{-21}}}\)
\(=1.264 \times 10^{-10} \mathrm{~m}\)
\(=1.264 \dot A\)

17.

\(Here,{ \phi }_{ 0 }=4.2eV\)
\(=4.2\times 1.6\times { 10 }^{ -19 }J,V=?,\lambda =2000\overset { \circ }{ A } =2000\times { 10 }^{ -10 }m\)
Max.K.E.of the emitted photoelectron.
\({ K }_{ max }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\( =\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 2000\times { 10 }^{ -10 } } -4.2\times 1.6\times { 10 }^{ -19 }J\)
\( =3.18\times { 10 }^{ -19 }J\)
\(Stopping \ potential,{ V }_{ 0 }=\frac { { K }_{ max } }{ e } =\frac { 3.18\times { 10 }^{ -19 } }{ 1.6\times { 10 }^{ -19 } } =1.9875V\)
(b) For threshold wavelength \({ \lambda }_{ 0 }\) the stopping potential is zero.
\({ \lambda }_{ 0 }=\frac { hc }{ { \phi }_{ 0 } } =\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 4.2\times 1.6\times { 10 }^{ -19 } } =2.946\times { 10 }^{ -7 }m=2946\overset { \circ }{ A } \)