Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 67

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Weekly Quiz Competition

Question 1
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What is the ratio of wavelength of a photon and that of an electron of mass, m of the same energy $$E$$?

A
$$C\sqrt {\dfrac {2m}{E}}$$

B
$$\sqrt {\dfrac {2m}{E}}$$

C
$$\dfrac{1}{C}\sqrt {\dfrac {2m}{E}}$$

D
$$\sqrt {\dfrac {m}{E}}$$

Solution

Wavelength of the photon is related to its energy by:
$$\lambda_p = \cfrac{hc}{E}$$

For an electron its deborglie wavelength is given by:
$$\lambda_e = \cfrac{h}{mv}$$
Squaring,
$$\lambda_e^2 = \cfrac{h^2}{m^2v^2}$$
$$= \cfrac{1/2h^2}{Em}$$
$$\lambda_e = \cfrac{h}{\sqrt{2Em}}$$

$$\cfrac{\lambda_p}{\lambda_e} = c \sqrt{ \cfrac{2m}{E} }$$
Question 2
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Calculate the momentum of particle whose de Broglie wavelength is $$2^oA$$.

A
$$3.313 \times 10^{-24}gms^{-1}$$

B
$$3.313 \times 10^{-24} kg m s^{-1}$$

C
$$33.13 \times 10^{-20}kgms^{-1}$$

D
none of these

Solution

De-broglie wavelength $$(\lambda) $$ = $$\frac{h}{mv} = \frac{h}{p} $$
Given , $$(\lambda) = 2 \times10^{-10} m$$
  $$p=\frac{h}{\lambda }$$ where h $$=6.626$$ x $$10^{-34}$$ Js.
  $$p=\frac{6.626x10^{-34}}{2x10^{-10}}kgm/s$$
  $$p=3.313x10^{-24}kgm/s$$
Hence, the momentum of a particle whose de-broglie wavelength is $$3.313\times10^{-24}kgm/s$$
Question 3
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An electron of mass 'm' has de-Broglie wavelength '$$\lambda$$' when accelerated through potential difference 'V'. When proton of mass 'M', is accelerated through potential difference $$9V$$, the de-Broglie wavelength associated with it will be : (Assume that wavelength is determined at low voltage)

A
$$\dfrac{\lambda}{3} \sqrt{\dfrac{M}{m}}$$

B
$$\dfrac{\lambda}{3} . \dfrac{M}{m}$$

C
$$\dfrac{\lambda}{3} \sqrt{\dfrac{m}{M}}$$

D
$$\dfrac{\lambda}{3} . \dfrac{m}{M}$$

Solution

De Broglie wavelength $$\lambda = \dfrac{h}{\sqrt{2mK}}$$

where $$K$$ is the kinetic energy.

Kinetic energy $$K =eV$$

We get $$\lambda = \dfrac{h}{\sqrt{2meV}}$$

$$\implies$$ $$\lambda \propto \dfrac{1}{\sqrt{mV}}$$ ........(1)

Given : $$V_1 = V$$ $$m_1 = m$$ $$m_2 = M$$ $$V_2 = 9V$$

From equation (1), we get $$\dfrac{\lambda_2}{\lambda_1} = \sqrt{\dfrac{m_1 V_1}{m_2V_2}}$$

Or $$\dfrac{\lambda_2}{\lambda} = \sqrt{\dfrac{m V}{M(9V)}}$$

$$\implies$$ $$\lambda_2 =\dfrac{\lambda}{3} \sqrt{\dfrac{m }{M}}$$
Question 4
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A proton when accelerated through a potential difference of $$V$$, has a de Broglie wavelength $$\lambda$$ associated with it. If an alpha particle is to have the same de Broglie wavelength $$\lambda$$, it must be accelerated through a potential difference of:

A
$$\displaystyle\frac{V}{8}$$

B
$$\displaystyle\frac{V}{4}$$

C
$$4V$$

D
$$8V$$

Solution

Since the kinetic energy will be equal to the work done on particle,

$$K=qV$$

Also, $$p=\sqrt { 2mK } $$

Thus, $$\lambda =\dfrac { h }{ p } =\dfrac { h }{ \sqrt { 2mK } } =\dfrac { h }{ \sqrt { 2mqV } } $$

For both to have same $$\lambda$$,

$${ m }_{ p }{ q }_{ p }{ V }={ m }_{ a }{ q }_{ a }{ V }_{ a }$$

$${ V }_{ a }=\dfrac { 1 }{ (4)(2) } V=\dfrac { V }{ 8 } $$

Answer is option A.
Question 5
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In hydrogen spectrum, the wavelength of the line is 656 nm, where as in the spectrum of a distant galaxy, the line wavelength is 706 nm. Estimated speed of galaxy with respect to the earth is :

A
$$ 2 \times 10^8 m/s $$

B
$$ 2 \times 10^7 m/s $$

C
$$ 2 \times 10^6 m/s $$

D
$$ 2 \times 10^5 m/s $$

Solution

$$ \dfrac { \triangle \lambda}{\lambda} = \pm \dfrac {v}{c} $$
Here (-) sign to be used for appraoching (+) sign to be used for receding.
$$ \dfrac { \triangle \lambda}{\lambda} = \dfrac {706-656}{656} = \pm = v/c $$
$$ \dfrac {v}{c} = \dfrac {50}{656} $$
$$ v = \dfrac {50}{656} \times c $$
$$ v = \dfrac {50}{656} \times 3 \times 10^8 = 2 \times10^7 m/s $$
Question 6
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An electronic transition from $$M$$ shell (n = 3) to $$K$$ shell (n =1) takes place in a hydrogenatom. Find the wavelength of radiation emited. ($$R =$$ 109, 677 $$cm^{-1}$$).

A
$$1026 A^o$$

B
$$1.026\times 10^{-5}m$$

C
$$1.026 \times 10^{-3}cm$$

D
none of these

Solution

For a hydrogen atom,
  $$\frac{1}{\lambda }=R\left [ \frac{1}{{n_{1}}^{2}}-\frac{1}{{n_{2}}^{2}} \right ]$$
Given $$n_{1}=1$$ , $$n_{2}=3$$
   $$\frac{1}{\lambda }=R\left [ 1-\frac{1}{9} \right ]$$
   $$\frac{1}{\lambda }=\frac{1}{R}\left [ \frac{9}{8} \right ]$$
$$=1.0257$$ x $$10^{-5}$$ cm
    $$=1025.7$$ x $$10^{-10}$$ m
$$\approx$$  $$1026$$ $$\mathring { A } $$
Question 7
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Figure represents a graph of kinetic energy of most energetic photoelectrons Km (in eV) and frequency $$\upsilon$$ for a metal used as cathode in photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is

A
$$1 \,\times \, 10^{14}Hz$$

B
$$1.5 \,\times \, 10^{14} Hz$$

C
$$2 \,\times \, 10^{14}Hz$$

D
$$2.7 \,\times \, 10^{14} Hz$$

Solution

From graph, $$\upsilon = 10^{15}\, Hz$$,
$$k_{max} \, = \, 3 eV \, = \, 3 \, \times \, 1.6 \, \times \, 10^{-19} \, J$$
According to Einsteins photoelectric equation
$$k_{max} \, = \, h\upsilon \, - \, h\upsilon_{0}$$
or $$h\upsilon_{0} \, = \, h\upsilon \, - \, k_{max}$$
or $$\upsilon_{0} \, = \, \upsilon \, - \dfrac{k_{max}}{h}$$ $$= \, 10^{15} \, - \, \dfrac{3 \, \times \, 1.6 \, \times \, 10^{-19}}{6.63 \, \times \, 10^{-34}}$$

$$\upsilon_{0} = \, (10 \, - \, 7.3) \, \times \, 10^{14} \, = \, 2.7 \, \times \, 10^{14} \, Hz$$
Question 8
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A particle A with a mass $$m_A$$ is moving with a velocity v and hits a particle B of mass $$m_B$$ at rest. If motion is one dimensional and take the collision is elastic, then the change in the de Broglie wavelength of the particle A is

A
$$\dfrac{h}{2m_A v} \left[\dfrac{(m_A \, + \, m_B)}{(m_A \, - \, m_B)} \, - \, 1\right]$$

B
$$\dfrac{h}{m_A v} \left[\dfrac{(m_A \, - \, m_B)}{(m_A \, + \, m_B)} \, - \, 1\right]$$

C
$$\dfrac{h}{m_A v} \left[\dfrac{(m_A \, + \, m_B)}{(m_A \, - \, m_B)} \, - \, 1\right]$$

D
$$\dfrac{2h}{m_A v} \left[\dfrac{(m_A \, + \, m_B)}{(m_A \, - \, m_B)} \, + \, 1\right]$$

Solution

According to law of conversation of momentum
$$m_Av \, + \, m_B \, \times \, 0 \, = \, m_Av_A \, + \, m_Bv_B$$
or $$m_A(v - v_B) \, = \, m_Bv_B$$ ....(i)
According to law of conversation of klinetic energy
$$\dfrac{1}{2}m_Av^2 \, = \, \dfrac{1}{2}m_Av^2_A \, + \, \dfrac{1}{2}m_Bv^2_B$$

or $$m_A(v^2 - V^2_A) = m_Bv^2_B$$

or $$m_A (v - v_A)(v + v_A) \, = \, m_Bv^2_b$$ ...(ii)

Dividing (ii) by (i), we get
$$v + v_A = v_B \, or \, v = v_B - v_A$$ ...(iii)

Solving (i) and (iii), we get
$$v_A = \left ( \dfrac{m_A - m_B}{m_A + m_B} \right )v \, $$ and $$ \, v_B = \left ( \dfrac{2m_A}{m_A + m_B} \right )v$$

$$\lambda_{initial} = \dfrac{h}{m_Av} $$ and $$ \, \, \lambda_{final} = \dfrac{h}{m_Av_A} = \dfrac{h(m_A + m_B)}{m_A(m_A - m_B)v}$$

$$\therefore \Delta\lambda = \lambda_{final} - \lambda_{initial} = \dfrac{h}{m_Av}\left [ \dfrac{(m_A + m_B)}{(m_A - m_B)} -1\right ]$$
Question 9
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Photons of wavelength $$\lambda$$ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius R by a perpendicular magnetic field having a magnitude B. The work function of the metal is (where symbols have their usual meanings).

A
$$\displaystyle\frac{hc}{\lambda}-m_e+\frac{e^2B^2R^2}{2m_e}$$

B
$$\displaystyle\frac{hc}{\lambda}+2m_e\left(\displaystyle\frac{eBR}{2m_e}\right)^2$$

C
$$\displaystyle\frac{hc}{\lambda}-m_ec^2-\frac{e^2B^2R^2}{2m_e}$$

D
$$\displaystyle\frac{hc}{\lambda}-2m_e\left(\displaystyle\frac{eBR}{2m_e}\right)^2$$

Solution
Question 10
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Photoelectron emission is observed for three different metals A, B and C. The kinetic energy of the fastest photoelectrons versus frequency 'V' is plouted for each metal . Which of the following graphs shows the phemomenon correctly.

A

B

C

D