Q: Photon of frequency 'v\ ' has a momentum associated with it. If 'c\ ' is the velocity of light, the momentum is

We know that energy of photon is E=h\nu ....(1) Momentum of photo , p=mc=mc^2/c=E/c where E=mc^2 using (1), p=h\nu/c

Q: Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are { \lambda }_{ 1 } and { \lambda }_{ 2 } , their de Broglie wavelength in the frame of reference attached to their centre of mass is:

Since \lambda = \dfrac{h}{mv} v = \dfrac{h}{m \lambda} Let v_1 and v_2 are the speeds of electrons v_{cm} = \dfrac{v_1 + v_2}{2} \dfrac{h}{2m \lambda_{cm}} = \dfrac{1}{2}( \dfrac{h}{m \lambda_1} + \dfrac{h}{m \lambda_2}) \dfrac{1}{\lambda_{cm}} = \dfrac{1}{\lambda_1} + \dfrac{1}{\lambda_2}

Q: The de-Broglie wavelength (\lambda_{B}) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (\lambda_{G}) by :

de- Broglie wavelength \lambda=\dfrac{h}{mv} By conservation of angular momentum, mvr=\dfrac{nh}{2\pi} \dfrac{h}{mv}=\dfrac{2\pi r}{n} \lambda=\dfrac{2\pi r}{n} r=a_{0}\dfrac{n^{2}}{Z} \lambda=\dfrac{2\pi a_{0}n}{Z} As the atom is hydrogen Z=1 \lambda_{B}=\dfrac{2\pi a_{0}3}{Z} \lambda_{G}=\dfrac{2\pi a_{0}}{Z} \lambda_{B}=3\lambda_{G}

Question 1

The de - Broglie wavelength associated with the electron in the  n=4  level is :

Accepted Answer

De-Broglie wavelength is given as  \lambda =\dfrac{h}{mv} ...(i) where h is planck constantThe velocity of electron in Bohr model is given as  v=\dfrac{h}{2\pi m_e a_\circ{} n} ...(i)in the ground state n=1 v_{n=1}=\dfrac{h}{2\pi m_e a_\circ{} }   v_{n=4}=\dfrac{h}{2\pi m_e a_\circ{} 4}   v_{n=4}=\dfrac{v_{n=1}}{4} ...(iii)from equation (i) \lambda_{n=4}=\dfrac{h}{m_e v_{n=4} } substituting value of  v_{n=4}  from equation (iii) \lambda_{n=4}=\dfrac{h}{m_e \dfrac{ v_{n=1}}{4} } \lambda_{n=4}=4 \lambda_{n=1} Hence correct answer will be option B.

Question 2

A particle  A  of mass  m  and initial velocity  v  collides with a particle  B  of mass  \dfrac{m}{2}  which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths  {\lambda}_A  to  {\lambda}_B  after the collision is :

Accepted Answer

From conservation of linear momentum,  mv= mv_B + \dfrac{m}{2} v_B v= v_A+ \dfrac{v_B}{2} Also   v^2 = v_a^2 + \dfrac{v_B^2}{2} {(v_A+ \dfrac{v_B}{2})}^2= v_A^2 + {(\dfrac{v_B}{2})}^2 v_Av_B = \dfrac{v_B}{4} v_B= 4 v_A  and  m_B= \dfrac{m_A}{2} P_A=m_Av_A P_B= \dfrac{m_A}{2} 	imes 4 v_A= 2 P_A \dfrac{\lambda_A}{\lambda_B}= \dfrac{P_B}{P_A}= \dfrac{2 P_A}{P_A}= 2

Question 3

For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?

Accepted Answer

A dust particle will it be most difficult to experimentally verify the de-Broglie relationship          .

Question 4

If electron charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental constant h are taken as fundamental quantities, the permeability of vacuum $μ_{0}$ can be expressed in units of

Accepted Answer

The fine structure constant is defined as  \alpha = \dfrac{e^2}{4\pi \epsilon _0 \hbar c}  and is dimensionless where  \hbar = \dfrac{h}{2\pi}  	herefore \alpha = \dfrac{e^2 c}{4\pi \epsilon _0 \hbar c^2}=\dfrac{e^2 c  \epsilon _0 \mu _0}{4\pi \epsilon _0 \hbar}  where  c = \dfrac{1}{\sqrt{\epsilon _0 \mu _0}}  \Rightarrow \mu _0 = \dfrac{ \hbar}{e^2 c}

Question 5

The potential energy of a particle varies as
$V (x) = E_{o}; 0 \leq x \leq 1$
$= 0; x > 1$
For $0 \leq x \leq 1$ , de Broglie wavelength is $γ_{1}$ and for $x > 1$ the de Broglie wavelength is $γ_{2}$ .
Total energy of the particle is $2 E_{o}$ . If $γ_{1} / γ_{2} = x$ . Find $x$

Accepted Answer

For  0\leq x\leq 1 , Kinetic energy  E_1=E_0=\dfrac{p_1^2}{2m} \Rightarrow p_1=\sqrt{2mE_0} For  x> 1 ,    E_2=2E_0=\dfrac{p_2^2}{2m} \Rightarrow p_2=\sqrt{4mE_0} now,  \gamma_1=\dfrac{h}{p_1}=\dfrac{h}{\sqrt{2mE_0}}   and  \gamma_2=\dfrac{h}{p_2}=\dfrac{h}{\sqrt{4mE_0}}  	herefore \dfrac{\gamma_1}{\gamma_2}=\dfrac{2}{\sqrt 2}=\sqrt 2

Question 6

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to:  (|e|=1.6 	imes 10^{-19} C, m_e=9.1 	imes 10^{-31}kg, h=6.6 	imes 10^{-34} Js).

Accepted Answer

\lambda = \dfrac{h}{p}= \dfrac{h}{\sqrt{2mE}}= \dfrac{h}{\sqrt{2mqV}} \lambda = \dfrac{ 6.6 	imes 10 ^ {-34} }{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 1.6 	imes 10^{-19} 	imes 50}} = 1.72  	imes 10^{-10} = 1.72 A  Option B

Question 7

Photon of frequency  'v\ '  has a momentum associated with it. If  'c\ '  is the velocity of light, the momentum is

Accepted Answer

We know that energy of photon is  E=h
u  ....(1)  Momentum of photo ,  p=mc=mc^2/c=E/c  where  E=mc^2 using (1),  p=h
u/c

Question 8

A parallel beam of electrons travelling in x-direction falls on a slit of width d. If after passing the slit, an electron acquires momentum  p_y  in the y direction, then for a majority of electrons passing through the slit (h is Planck's constant).

Accepted Answer

From Heisenberg's uncertainty principle,   \Delta p \Delta y \geq h For most electrons,  \Delta p < p_y  (change in momentum is small )                           and  \Delta y < d  (most electrons lie in central fringe)  \Rightarrow |p_y|d \geq h Minimum value of   \Rightarrow |p_y|d  is  h .

Question 9

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are  { \lambda  }_{ 1 }  and  { \lambda  }_{ 2 } , their de Broglie wavelength in the frame of reference attached to their centre of mass is:

Accepted Answer

Since  \lambda = \dfrac{h}{mv} v = \dfrac{h}{m \lambda} Let  v_1  and  v_2  are the speeds of electrons v_{cm} = \dfrac{v_1 + v_2}{2} \dfrac{h}{2m \lambda_{cm}} = \dfrac{1}{2}( \dfrac{h}{m \lambda_1} + \dfrac{h}{m \lambda_2}) \dfrac{1}{\lambda_{cm}} = \dfrac{1}{\lambda_1} + \dfrac{1}{\lambda_2}

Question 10

The de-Broglie wavelength  (\lambda_{B})  associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state  (\lambda_{G})  by :

Accepted Answer

de- Broglie wavelength  \lambda=\dfrac{h}{mv} By conservation of angular momentum,  mvr=\dfrac{nh}{2\pi} \dfrac{h}{mv}=\dfrac{2\pi r}{n} \lambda=\dfrac{2\pi r}{n} r=a_{0}\dfrac{n^{2}}{Z} \lambda=\dfrac{2\pi a_{0}n}{Z} As the atom is hydrogen  Z=1 \lambda_{B}=\dfrac{2\pi a_{0}3}{Z} \lambda_{G}=\dfrac{2\pi a_{0}}{Z} \lambda_{B}=3\lambda_{G}

Dual Nature Of Radiation And Matter

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The de - Broglie wavelength associated with the electron in the $n = 4$ level is :

A particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $λ_{A}$ to $λ_{B}$ after the collision is :

For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?

If electron charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental constant h are taken as fundamental quantities, the permeability of vacuum $μ_{0}$ can be expressed in units of

The potential energy of a particle varies as
$V (x) = E_{o}; 0 \leq x \leq 1$
$= 0; x > 1$
For $0 \leq x \leq 1$ , de Broglie wavelength is $γ_{1}$ and for $x > 1$ the de Broglie wavelength is $γ_{2}$ .
Total energy of the particle is $2 E_{o}$ . If $γ_{1} / γ_{2} = x$ . Find $x$

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: $(∣ e ∣ = 1.6 \times 1 0^{- 19} C, m_{e} = 9.1 \times 1 0^{- 31} k g, h = 6.6 \times 1 0^{- 34} J s) .$

Photon of frequency $^{'} v^{'}$ has a momentum associated with it. If $^{'} c^{'}$ is the velocity of light, the momentum is

A parallel beam of electrons travelling in x-direction falls on a slit of width d. If after passing the slit, an electron acquires momentum $p_{y}$ in the y direction, then for a majority of electrons passing through the slit (h is Planck's constant).

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $λ_{1}$ and $λ_{2}$ , their de Broglie wavelength in the frame of reference attached to their centre of mass is:

The de-Broglie wavelength $(λ_{B})$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $(λ_{G})$ by :

Dual Nature Of Radiation And Matter

Physics 0 View

The de - Broglie wavelength associated with the electron in the n=4 level is :

A particle A of mass m and initial velocity v collides with a particle B of mass 2m​ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA​ to λB​ after the collision is :

For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?

If electron charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental constant h are taken as fundamental quantities, the permeability of vacuum μ0​ can be expressed in units of

The potential energy of a particle varies as V(x)=Eo​;0≤x≤1 =0;x>1 For 0≤x≤1, de Broglie wavelength is γ1​ and for x>1 the de Broglie wavelength is γ2​. Total energy of the particle is 2Eo​. If γ1​/γ2​=x​. Find x

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: (∣e∣=1.6×10−19C,me​=9.1×10−31kg,h=6.6×10−34Js).

Photon of frequency ′v ′ has a momentum associated with it. If ′c ′ is the velocity of light, the momentum is

A parallel beam of electrons travelling in x-direction falls on a slit of width d. If after passing the slit, an electron acquires momentum py​ in the y direction, then for a majority of electrons passing through the slit (h is Planck's constant).

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ1​ and λ2​, their de Broglie wavelength in the frame of reference attached to their centre of mass is:

The de-Broglie wavelength (λB​) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (λG​) by :

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The de - Broglie wavelength associated with the electron in the $n = 4$ level is :

A particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $λ_{A}$ to $λ_{B}$ after the collision is :

If electron charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental constant h are taken as fundamental quantities, the permeability of vacuum $μ_{0}$ can be expressed in units of

The potential energy of a particle varies as
$V (x) = E_{o}; 0 \leq x \leq 1$
$= 0; x > 1$
For $0 \leq x \leq 1$ , de Broglie wavelength is $γ_{1}$ and for $x > 1$ the de Broglie wavelength is $γ_{2}$ .
Total energy of the particle is $2 E_{o}$ . If $γ_{1} / γ_{2} = x$ . Find $x$

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: $(∣ e ∣ = 1.6 \times 1 0^{- 19} C, m_{e} = 9.1 \times 1 0^{- 31} k g, h = 6.6 \times 1 0^{- 34} J s) .$

Photon of frequency $^{'} v^{'}$ has a momentum associated with it. If $^{'} c^{'}$ is the velocity of light, the momentum is

A parallel beam of electrons travelling in x-direction falls on a slit of width d. If after passing the slit, an electron acquires momentum $p_{y}$ in the y direction, then for a majority of electrons passing through the slit (h is Planck's constant).

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $λ_{1}$ and $λ_{2}$ , their de Broglie wavelength in the frame of reference attached to their centre of mass is:

The de-Broglie wavelength $(λ_{B})$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $(λ_{G})$ by :