Physics

9.4 Read the following two statements below carefully and state, with reasons, if it is true or false.

The Young's modulus of rubber is greater than that of steel

The stretching of a coil is determined by its shear modulus

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For a given stress, the strain in rubber is more than it is in steel, hence the Young's modulus of rubber is lesser than in steel. So the statement is False.

Shear modulus is the ratio of the applied stress to the change in the shape of a body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

= 2.2 $* 10^{- 4} m^{3}$

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9.3 The stress-strain graphs for materials A and B are shown in Fig. 9.12.

The graphs are drawn to the same scale.

(a) Which of the materials has the greater Young's modulus?

(b) Which of the two is the stronger material?

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Material A has greater Young's modulus.

Material A is the strongest as it can withstand more strain than material B without fracture.

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9.2 Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?

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From the given graph, for the value stress 150 $* 10^{6}$ N/ $m^{2}$ , the strain is 0.002

Young's modulus = $\frac{150 * 10^{6}}{0.002}$ = 7.5 $* 10^{10}$ N/ $m^{2}$

Yield strength is the maxium strength the material can withstand in elastic limit. From the graph, the yield strength is 300 $* 10^{6}$ or 3 $* 10^{8} N / m^{2}$

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9.1 A steel wire of length 4.7 m and cross-sectional area 3.0 * 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 * 10^–5 m² under a given load. What is the ratio of the Young's modulus of steel to that of copper?

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Length of the steel wire, $L_{1}$ = 4.7 m

Area of cross-section of the steel wire, $A_{1}$ = 3.0 $* 10^{- 5}$ m2

Length of the copper wire, $L_{2}$ = 3.5 m

Area of cross-section of the copper wire, $A_{2}$ = 4.0 $* 10^{- 5}$ m2

Change in length, $Δ {L =}_{} L_{1} - L_{2}$

Let the force applied = F

Young's modulus in steel wire,

$Y_{1}$ = $\frac{F_{1}}{A_{1}} * \frac{L_{1}}{Δ L_{}}$ ….(1)

Young's modulus in copper wire,

$Y_{2}$ = $\frac{F_{2}}{A_{2}} * \frac{L_{2}}{Δ L_{}}$ …….(2)

The ratio of Young's modulus

$\frac{Y_{1}}{Y_{2}}$ = $\frac{F_{1}}{A_{1}} * \frac{L_{1}}{Δ L_{}} * \frac{A_{2}}{F_{2}} * \frac{Δ L_{}}{L_{2}}$ = $\frac{L_{1}}{A_{1}} * \frac{A_{2}}{L_{2}}$ = $\frac{4.7 * 4 * 10^{- 5}}{3 * 10^{- 5} * 3.5}$ = $\frac{18.8}{10.5} = 1.79$

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11.20 (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 * 10¹¹ C kg^–1.

(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified?

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11.20 Potential difference across evacuated tube, V = 500 V

Specific charge of electron, e/m = 1.76 $* 10^{11}$ C/kg

The speed of each emitted electron is given by the relation of kinetic energy as

$E_{k i n e t i c}$ = $\frac{1}{2} m v^{2}$ = $e V$

$v = \sqrt{\frac{2 e V}{m}}$ = $\sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 * 500 * 1.76 * 10^{11}}$ = 13.27 $* 10^{6}$ m/s

Therefore, the speed of each emitted electron is 13.27 $* 10^{6}$ m/s

Collector potential, V = 10 MV = 10 $* 10^{6}$ V

The speed is given by $v = \sqrt{\frac{2 e V}{m}}$ $= \sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 * 10 * 10^{6} * 1.76 * 10^{11}}$ = 1.876 $* 10^{9}$ m/s

This is not possible nothing can move faster than the light. In the above formula

$E_{k i n e t i c}$ = $\frac{1}{2} m v^{2}$ can only be used in the non-relativistic limit, i.e. v << c

For very high speed problems

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11.20 Potential difference across evacuated tube, V = 500 V

Specific charge of electron, e/m = 1.76 $* 10^{11}$ C/kg

The speed of each emitted electron is given by the relation of kinetic energy as

$E_{k i n e t i c}$ = $\frac{1}{2} m v^{2}$ = $e V$

$v = \sqrt{\frac{2 e V}{m}}$ = $\sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 * 500 * 1.76 * 10^{11}}$ = 13.27 $* 10^{6}$ m/s

Therefore, the speed of each emitted electron is 13.27 $* 10^{6}$ m/s

Collector potential, V = 10 MV = 10 $* 10^{6}$ V

The speed is given by $v = \sqrt{\frac{2 e V}{m}}$ $= \sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 * 10 * 10^{6} * 1.76 * 10^{11}}$ = 1.876 $* 10^{9}$ m/s

This is not possible nothing can move faster than the light. In the above formula

$E_{k i n e t i c}$ = $\frac{1}{2} m v^{2}$ can only be used in the non-relativistic limit, i.e. v << c

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as :

$E = m c^{2}$ , where m = relativistic mass

$E = m_{0} \sqrt{(1 - \frac{v^{2}}{c^{2}}})$ where $m_{0}$ = Mass of the particle at rest.

Hence, kinetic energy is given as

$E_{k i n e t i c}$ = m $c^{2}$ - $m_{0} c^{2}$

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11.19 What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

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11.19 Temperature of the Nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of Nitrogen molecule, m = 2 $*$ 14.0076 u = 28.0152 u

We know, 1 u = 1.66 $* 10^{- 27}$ kg

So, m = 28.0152 $*$ 1.66 $* 10^{- 27}$ kg = 4.65 $* 10^{- 26}$ kg

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

Boltzmann constant, k = 1.38 $* 10^{- 23}$ kg $m^{2} s^{- 2} K^{- 1}$

We have the expression that relates to mean kinetic energy ( $\frac{3}{2} k T)$ of the nitrogen molecule with root mean square speed ( $v_{r m s}$ ) as:

$\frac{1}{2} m {v_{r m s}}^{2}$ = $\frac{3}{2} k T$

$v_{r m s} = \sqrt{\frac{3 k T}{m}}$ = $\sqrt{\frac{3 * 1.38 * 10^{- 23} * 300}{4.65 * 10^{- 26}}}$ = 516.814 m/s

De Broglie wavelength of the nitrogen molecule

$λ = \frac{h}{m v_{r m s}}$ = $\frac{6.626 * 10^{- 34}}{4.65 * 10^{- 26} * 516.814}$ = 2.76 $* 10^{- 11}$ m = 0.0276 nm

Therefore,

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11.19 Temperature of the Nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of Nitrogen molecule, m = 2 $*$ 14.0076 u = 28.0152 u

We know, 1 u = 1.66 $* 10^{- 27}$ kg

So, m = 28.0152 $*$ 1.66 $* 10^{- 27}$ kg = 4.65 $* 10^{- 26}$ kg

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

Boltzmann constant, k = 1.38 $* 10^{- 23}$ kg $m^{2} s^{- 2} K^{- 1}$

We have the expression that relates to mean kinetic energy ( $\frac{3}{2} k T)$ of the nitrogen molecule with root mean square speed ( $v_{r m s}$ ) as:

$\frac{1}{2} m {v_{r m s}}^{2}$ = $\frac{3}{2} k T$

$v_{r m s} = \sqrt{\frac{3 k T}{m}}$ = $\sqrt{\frac{3 * 1.38 * 10^{- 23} * 300}{4.65 * 10^{- 26}}}$ = 516.814 m/s

De Broglie wavelength of the nitrogen molecule

$λ = \frac{h}{m v_{r m s}}$ = $\frac{6.626 * 10^{- 34}}{4.65 * 10^{- 26} * 516.814}$ = 2.76 $* 10^{- 11}$ m = 0.0276 nm

Therefore, De Broglie wavelength is 0.0276 nm.

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11.18 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

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11.18 The momentum of a photon having energy ( $h ν)$ is given as $p$ = $\frac{h ν}{c}$ = $\frac{h}{λ}$

So, $λ = \frac{h}{p}$ …………….(1)

Where, $λ =$ Wavelength of the electromagnetic radiation

h = Planck's constant

c = speed of light

De Broglie wavelength of the photon is given as $λ = \frac{h}{m v}$

But momentum, $= m v$ , where $m$ = mass of the photon, $v$ = velocity of the photon

Hence $λ = \frac{h}{p}$ ……….(2)

Hence, it can be inferred from equation (1) and (2) that wavelength of the electromagnetic radiation is equal to the De Broglie wavelength of the photon.

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11.17 (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 * 10^–10 m?

(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.

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11.17 De Broglie wavelength of the neutron, $λ$ = 1.40 $* 10^{- 10}$ m

Mass of neutron, m = 1.66 $* 10^{- 27}$ kg

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

Kinetic energy, $E_{k}$ = $\frac{1}{2} m v^{2}$ ………….(1)

De Broglie wavelength and velocity (v) are related as

$λ = \frac{h}{m v}$ …………….(2)

Combining equation (1) and (2), we get

$E_{k}$ = $\frac{1}{2} m ({\frac{h}{m λ})}^{2}$ = $\frac{h^{2}}{2 m λ^{2}}$ = $\frac{{(6.626 * 10^{- 34})}^{2}}{2 * 1.66 * 10^{- 27} * ({1.40 * 10^{- 10})}^{2}}$ = 6.75 $* 10^{- 21}$ J = $\frac{6.75 * 10^{- 21}}{1.6 * 10^{- 19}}$ eV = 42.17 $* 10^{- 3} e V$

Temperature of the neutron, T = 300 K

Average kinetic energy of the neutron, $E_{k - a v g}$ = $\frac{3}{2} k T$ ,

where k = Boltzmann constant = 1.38 $* 10^{- 23}$ kg $m^{2} s^{- 2} K^{- 1}$

$E_{k - a v g}$ = $\frac{3}{2} k T$ = $\frac{3}{2} * 1.38 * 10^{- 23} * 300 = 6.21 * 10^{- 21}$ J

The relationship of De Broglie wa

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11.17 De Broglie wavelength of the neutron, $λ$ = 1.40 $* 10^{- 10}$ m

Mass of neutron, m = 1.66 $* 10^{- 27}$ kg

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

Kinetic energy, $E_{k}$ = $\frac{1}{2} m v^{2}$ ………….(1)

De Broglie wavelength and velocity (v) are related as

$λ = \frac{h}{m v}$ …………….(2)

Combining equation (1) and (2), we get

Temperature of the neutron, T = 300 K

Average kinetic energy of the neutron, $E_{k - a v g}$ = $\frac{3}{2} k T$ ,

where k = Boltzmann constant = 1.38 $* 10^{- 23}$ kg $m^{2} s^{- 2} K^{- 1}$

$E_{k - a v g}$ = $\frac{3}{2} k T$ = $\frac{3}{2} * 1.38 * 10^{- 23} * 300 = 6.21 * 10^{- 21}$ J

The relationship of De Broglie wavelength is given as:

$λ = \frac{h}{\sqrt{2 * E_{k - a v g} * m}}$ = $\frac{6.626 * 10^{- 34}}{\sqrt{2 * 6.21 * 10^{- 21} * 1.66 * 10^{- 27}}}$ = 1.46 $* 10^{- 10}$ m = 0.146 nm

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11.16 An electron and a photon each have a wavelength of 1.00 nm. Find

(a) Their momentum,

(b) The energy of the photon, and

(c) The kinetic energy of electron.

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11.16 Wavelength of electron, $λ_{e}$ = Wavelength of proton, $λ_{p}$ = 1.0 nm = 1 $* 10^{- 9}$ m

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

From De Broglie wavelength relation, $λ = \frac{h}{p},$ where p = momentum

$p$ = $\frac{h}{λ}$ = $\frac{6.626 * 10^{- 34}}{1 * 10^{- 9}}$ = 6.626 $* 10^{- 25}$ kg.m/s. Since $λ_{e} = λ_{p}$ , their momentum will be also equal.

The energy of photon is given by the relation: $E$ = $\frac{h c}{λ},$

where c = speed of light = 3 $* 10^{8}$ m/s

$E$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{1 * 10^{- 9}}$ Js = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{1 * 10^{- 9} * 1.6 * 10^{- 19}}$ eV = 1242.38 eV = 1.242 keV

Kinetic energy of electron, having momentum p is given by the relation

$E_{k}$ = $\frac{1}{2} \frac{p^{2}}{m}$ where m = mass of electron = 9.1 $* 10^{- 31}$ kg.

Hence, $E_{k}$ &nbs

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11.16 Wavelength of electron, $λ_{e}$ = Wavelength of proton, $λ_{p}$ = 1.0 nm = 1 $* 10^{- 9}$ m

Planck's constant, h = 6.626 $* 10^{- 34}$ Js

From De Broglie wavelength relation, $λ = \frac{h}{p},$ where p = momentum

$p$ = $\frac{h}{λ}$ = $\frac{6.626 * 10^{- 34}}{1 * 10^{- 9}}$ = 6.626 $* 10^{- 25}$ kg.m/s. Since $λ_{e} = λ_{p}$ , their momentum will be also equal.

The energy of photon is given by the relation: $E$ = $\frac{h c}{λ},$

where c = speed of light = 3 $* 10^{8}$ m/s

$E$ = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{1 * 10^{- 9}}$ Js = $\frac{6.626 * 10^{- 34} * 3 * 10^{8}}{1 * 10^{- 9} * 1.6 * 10^{- 19}}$ eV = 1242.38 eV = 1.242 keV

Kinetic energy of electron, having momentum p is given by the relation

$E_{k}$ = $\frac{1}{2} \frac{p^{2}}{m}$ where m = mass of electron = 9.1 $* 10^{- 31}$ kg.

Hence, $E_{k}$ = $\frac{1}{2} \frac{{(6.626 * 10^{- 25})}^{2}}{9.1 * 10^{- 31}}$ J = 2.412 $* 10^{- 19}$ J = $\frac{2.412 * 10^{- 19}}{1.6 * 10^{- 19}}$ eV = 1.51 eV

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11.15 What is the de Broglie wavelength of

(a) Bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,

(b) A ball of mass 0.060 kg moving at a speed of 1.0 m/s, and

(c) A dust particle of mass 1.0 * 10^–9 kg drifting with a speed of 2.2 m/s?

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