Question Set B

CBSE 12th Physics Atoms Important Questions And Answers Study Material - QB365

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Questions:

3 Marks

What is the de-Broglie wavelength of a nitrogen molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

The electric field associated with a monochromatic beam of light becomes zero, \(2.4\times { 10 }^{ 15 }\) times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0eV, h=\(6.63\times { 10 }^{ -34 }Js\)

The work function for cesium is 1.8eV. Light of \(4500\mathring { A } \) is incident on it. Calculate
(i) the maximum kinetic energy of the emitted photoelectron
(ii) maximum velocity of the emitted photoelectron
(iii) if the intensity of the incident light is doubled, then find the maximum kinetic energy of the emitted photoelectron
Given \(h=6.6\times { 10 }^{ -34 }Js,\ { m }_{ e }=9.1\times { 10 }^{ -31 }kg, \ c=3\times { 10 }^{ 8 }{ ms }^{ -1 }\)

A photon of wavelength \(3310\mathring { A } \)falls on a photocathode and an electron of energy \(3\times { 10 }^{ -19 }\)J is ejected. If the wavelength of the incident photon is changed to \(5000\mathring { A } \), the energy of the ejected electron is \(9.72\times { 10 }^{ -20 }\)J. Calculate the value of Planck's constant and threshold wavelength of the photon.

A neutron strikes \(_{ 5 }{ { B }^{ 10 } }\)nucleus with the subsequent emission of an alpha particle. What is the atomic number, mass number and chemical name of the remaining nucleus?

If each diode in figure has a forward bias resistance of \(25\Omega \) and infinite resistance in reverse bias, what will be the values of the currents \(I_{ 1 },I_{ 2 },I_{ 3 } \ and \ I_{ 4 }\) ?

Assuming an ideal diode, draw the output waveform for the circuit given in the figure, explain the waveform.

(i) In the following diagram, which bulb out of B₁ and B₂ will glow and why?

(ii) Draw a diagram of an illuminated p-n. junction solar cell.
(iii) Explain briefly the three processes due to which generation of emf takes place in a solar cell.

10.

In half wave rectification , what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full wave rectification for the same input frequency.

11.

Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of \({ }_{8}^{16} \mathrm{O} \text { in } \mathrm{MeV} / \mathrm{c}^{2}\)

12.

From the relation R = R₀A^1/3, where R₀ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Case Study Questions

13.

From Bohr's atomic model, we know that the electrons have well defined energy levels in an isolated atom. But due to interatomic interactions in a crystal, the electrons of the outer shells are forced to have energies different from those in isolated atoms. Each energy level splits into a number of energy levels forming a continuous band.The gap between top of valence band and bottom of the conduction band in which no allowed energy levels for electrons can exist is called energy gap.

(i) In an insulator energy band gap is

(a) E_g = 0

(b) E_g< 3eV

(c) E_g> 3eV

(d) None of the above

(ii) In a semiconductor, separation between conduction and valence band is of the order of

(a) 0 eV

(b) 1 eV

(c) 10 eV

(d) 50 eV

(iii) Based on the band theory of conductors, insulators and semiconductors, the forbidden gap is smallest in

(a) conductors

(b) insulators

(c) semiconductors

(d) All of these

(iv) Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate?

(a) The number of free electrons for conduction is significant only in Si and Ge but small in C.

(b) The number of free conduction electrons is significant in C but small in Si and Ge.

(c) The number of free conduction electrons is negligibly small in all the three.

(d) The number offree electrons for conduction is significant in all the three.

(v) Solids having highest energy level partially filled with electrons are

(a) semiconductor

(b) conductor

(c) insulator

(d) none of these

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Answers Key:

3 Marks

Given mass of nitrogen molecule
= 2 x 14.0076 u
= \(28.0152\times1.67\times{ 10 }^{ -27 }kg\)
\(T=300k\)
Since de-Broglie wavelength.
\(\lambda =\frac { h }{ \sqrt { 3mkT } } \)
\(\lambda =\frac { 6.62\times{ 10 }^{ -34 } }{ \sqrt { 3\times28.0152\times1.67\times{ 10 }^{ -27 }\times1.38\times{ 10 }^{ -23 }\times300 } } \)
\(=\frac { 6.62\times{ 10 }^{ -34 } }{ 241 } \)
= \(0.0275\times{ 10 }^{ -9 }\)
= 0.028 nm

If r₀ is Bohr radius and v₀ is the velocity of electron in 1st orbit, then
Time period \(T={2\pi r_0\over v_0}\)
So \(I={e\over T}={ev_0\over2\pi r_0}\)

Given, \(\phi_{0}=2.0 \mathrm{eV} ; h-6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}, \mathrm{KE}_{\max }-?\)
In one complete vibration twice the electric field becomes zero, so the frequency of incident light is given by
\(\mathrm{V}=\frac{1}{2} \times 2.4 \times 10^{15}=1.2 \times 10^{15} \mathrm{~Hz}\)
Hence, maximum kinetic energy,
\(\mathrm{KE}_{\max }=h v-\phi_{0}=\frac{6.63 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}}-2=2.97 \mathrm{eV}\)

\(Here \ { \phi }_{ 0 }=1.8eV,\lambda =4500 \ A=4.5\times { 10 }^{ -7 }m\)
(i) Max K.E of emitted photoelectron is
\({ K }_{ max }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\( =\frac { (6.6\times { 10 }^{ -34 })(3\times { 10 }^{ 8 }) }{ 4.5\times { 10 }^{ -7 } } -1.8\times 1.6\times { 10 }^{ -19 }\)
\( =4.4\times { 10 }^{ -19 }-2.88\times { 10 }^{ -19 }=1.52\times { 10 }^{ -19 }J\)
(ii)Max. velocity of emitted photoelectron
\({ v }_{ max }=\sqrt { \frac { { 2K }_{ max } }{ m } } =\sqrt { \frac { 2\times 1.52\times { 10 }^{ -19 } }{ 9.1\times { 10 }^{ -31 } } }\)
\( =5.78\times { 10 }^{ 5 }{ ms }^{ -1 }\)
(iii) The kinetic energy of the emitted photoelectron is an incident of the intensity of the incident light. Hence, if the intensity of incident light is doubled the max. K.E of the emitted photoelectron electrons remains unchanged.

\(6.62\times { 10 }^{ -34 }Js\)
\(6620\mathring { A } \)

\(_{ 3 }{ { Li }^{ 7 } }\)

Given, forward biased resistance = \(25\Omega \)
Reverse biased resistance = \(\infty \)
As the diode in branch CD is in reverse biased which having resistance infinite.
so \(I_{ 3 }\)= 0
Resistance in branch AB = 25 + 125 = 150\(\Omega \)(say R₁)
Resistance in branch EF = 25 + 125 = 150\(\Omega \)(say R₂)
AB is parallel to EF.
so, resultant resistance, \(\frac { 1 }{ { R }^{ \prime } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 1 }{ 150 } +\frac { 1 }{ 150 } =\frac { 2 }{ 150 } \)
\(\Rightarrow { R }^{ \prime }=75\Omega \)
Total resistance, \(R={ R }^{ \prime }+25=75+25=100\Omega \)
Current, \({ I }_{ 1 }=\frac { V }{ R } =\frac { 5 }{ 100 } =0.05A\)
\({ I }_{ 1 }={ I }_{ 4 }+{ I }_{ 2 }+{ I }_{ 3 }\quad [Here,\quad { I }_{ 3 }=0]\)
\(\\ { I }_{ 1 }={ I }_{ 4 }+{ I }_{ 2 }\)
Here, the resistances R₁ and R₂ are same
\({ I }_{ 4 }={ I }_{ 2 }\)
\(\\ { I }_{ 1 }=2{ I }_{ 2 }\)
\(\\ \Rightarrow { I }_{ 2 }=\frac { { I }_{ 1 } }{ 2 } =\frac { 0.05 }{ 2 } =0.025A\)
\(\\ { I }_{ 4 }=0.025A\)
\(\\ { I }_{ 1 }=0.05A\)
\(\\ { I }_{ 2 }=0.025A\)
\(\\ { I }_{ 3 }=0\)
\(\\ { I }_{ 4 }=0.025A\)

When the input voltage is equal to or less than 5 V, diode will be reverse biased. It will offer high resistance in comparison to resistance (R) in series. now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sine wave input is to dip off all positive going portion above 5 V.
If the input voltage is more than +5V, the diode will be conducting as forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the voltage beyond +5V will appear across R.
When input voltage is negative, input voltage becomes more than -5 V (due to 5V battery in reverse bias position), the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now, junction diode appears in open circuit. The input waveform is then passed on to the output terminals. The output waveform is shown here in the figure

(i) D₁ diode is forward biased, hence current will flow in B₁ bulb and D₂ is reverse biased, so there will be no current in B2. Hence, BI will glow.
(ii) The diagram of illuminated p-n junction solar cell is given below

(iii) Processes due to generation of emf takes place in a solar cell are given below
(a) When light photon reach the junction, the excited electrons from the valence band to conduction band creating equal number of holes and electrons.
(b) These electron hole pair move in opposite direction due to junction field. Their movement in opposite direction creates potential difference (photo-voltage).
(c) When load is connected in the external circuit, current starts flowing through it due to photo-voltage.

10.

Given, input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
\(\therefore\) Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
\(\therefore\) Output frequency = 2 x 50 = 100 Hz.

11.

1u = 1.6605 x 10^–27 kg
To convert it into energy units, we multiply it by c² and find that energy equivalent = \(1.6605 \times 10^{-27} \times\left(2.9979 \times 10^{8}\right)^{2} \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}^{2}\)
\(=1.4924 \times 10^{-10} \mathrm{~J}\)
\(=\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}} \mathrm{eV}\)
\(=0.9315 \times 10^{9} \mathrm{eV}\)
\(=931.5 \mathrm{MeV}\)
or, \(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
For, \({ }_{8}^{16} \mathrm{O}, \quad \Delta M=0.13691 \mathrm{u}=0.13691 \times 931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
\(=127.5 \mathrm{MeV} / \mathrm{c}^{2}\)
The energy needed to separate \({ }_{8}^{16} \mathrm{O}\) into its constituents is thus 127.5 MeV/c².

12.

We have the expression for nuclear radius as:
R = R₀A¹/³
Where,
R₀ = Constant.
A = Mass number of the nucleus
Nuclear matter density, \(\rho \ =\frac{Mass\ of\ the\ nucles}{Volume\ of\ the\ nucles}\)
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
\(\therefore \rho=m \frac{A}{\frac{4}{3} \pi R^{3}}=\frac{3 \mathrm{~mA}}{4 \pi\left(R_{0} A^{\frac{1}{3}}\right)^{3}}=\frac{3 m A}{4 \pi R_{0}^{3} A}=\frac{3 \mathrm{~m}}{4 \pi R_{0}^{3}}\)
Hence, the nuclear matter density is independent of A. It is nearly constant.

Case Study Questions

13.

(i) (c) :In insulator, energy band gap is > 3 eV
(ii) (b) : In conductor, separation between conduction and valence bands is zero and in insulator, it is greater than 1 eV. Hence in semiconductor the separation between conduction and valence band is 1 eV.
(iii) (a): According to band theory the forbidden gap in conductors E_g = 0, in insulators E_g > 3 eV and in semiconductors E_g < 3 eV.
(iv) (a): The four valence electrons of C, Si and Ge lie respectively in the second, third and fourth orbit.Hence energy required to take out an electron from these atoms (i.e. ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, the number of free electrons for conduction in Ge and Si are significant but negligibly small for C.
(v) (b)