Q: An electron beam after collision with the target produces X-rays of wavelength \$4 A^0\$. The velocity of the electron beam is

Energy of x-ray beam \$= \frac{hc}{ \lambda} = 4.95*10^{-16} J =\frac{1}{2} mv^2\$ and it gives \$v = 6.31*10^7\$ m/s.So, the answer is option (B).

Q: A proton when accelerated through a potential difference of V volt has a wavelength \$\lambda \$ associated with it. An \$\alpha \$ - particle in order to have the same wavelength \$\lambda \$ must be accelerated through a p.d. of

For proton, \$\lambda_p =\dfrac{0.286\times 10^{-10}}{\sqrt{\nu }}\ m\$where V is p.dFor \$\alpha -particle,\ \lambda_\alpha =\dfrac{0.101\times 10^{-10}}{\sqrt{\nu }}m\$so, keeping same \$\alpha \$ for proton and \$\alpha -partside\$\$1= \dfrac{\lambda _p}{\lambda _{\alpha }} = \dfrac{0.286 \times \sqrt{V_{\alpha }}}{0.101 \times \sqrt{V_{P}}}\$\$0.101 \times \sqrt{V_{P}} = 0.286 \times \sqrt{V_{\alpha }}\$\$(0.101)^{2} \times V_{P} = (0.286)^{2}V _{\alpha} \$\$\dfrac{(0.101)^{2}\nu }{(0.286)^{2}} = V _{\alpha }\$ \$(\because V_{P}=V)\$\$\Rightarrow V_{\alpha } = \dfrac{V}{8} volt\$So, the answer is option (A).

Q: The momentum of a proton is p. The corresponding wavelength is

\$\lambda =\dfrac{h}{mv}\$ \$=\dfrac{h}{p}\$

Q: Moving with the same velocity, one of the following has the longest de Broglie wavelength

De-broglie wavelength\$\lambda =\dfrac{h}{mv}\$\$\lambda \propto \dfrac{1}{m}\$\$\beta\$ particle has least mass.So \$\beta\$ particle has the longest wavelength.

Q: Einstein’s photoelectric equation states that\$E_{k} = h v - W\$, In this equation \$E_k\$ refers to :

\$E_{k}=h\nu -\phi \$ (here \$\phi=W\$)here \$E_{k}\$ refers to the maximum kinetic energy of emitted electrons.

Q: The wavelength of matter waves does not depend on

Wavelength of matter,\$\lambda =\dfrac{h}{mv}\$

Q: De-Broglie wavelength depends on

\$\lambda =\dfrac{h}{mv}\$So \$\lambda \$ depends on mass of the particle.

Q: The Einstein’s photo-electric equation is based upon the conservation of

Einsteins photoelectric equations states that\$E_{k}=h\nu - \phi \$where, \$E_{k}\$ is maximum kinetic energy of emitted photoelectronsh is plank's constant.\$\nu \$ is frequency of incident light\$\phi \$ is work function of metal.So, it is clearly based upon energy conservation.

Question 1

An electron beam after collision with the target produces X-rays of wavelength $4 A^0$. The velocity of the electron beam is

Accepted Answer

Energy of x-ray beam $= \frac{hc}{ \lambda} = 4.95*10^{-16} J =\frac{1}{2} mv^2$ and it gives $v = 6.31*10^7$ m/s.So, the answer is option (B).

Question 2

A proton when accelerated through a potential difference of V volt has a wavelength $λ$ associated with it. An $α$ - particle in order to have the same wavelength $λ$ must be accelerated through a p.d. of

Accepted Answer

For proton, $\lambda_p =\dfrac{0.286\times 10^{-10}}{\sqrt{\nu }}\ m$where V is p.dFor $\alpha -particle,\ \lambda_\alpha =\dfrac{0.101\times 10^{-10}}{\sqrt{\nu }}m$so, keeping same $\alpha $ for proton and $\alpha -partside$$1= \dfrac{\lambda _p}{\lambda _{\alpha }} = \dfrac{0.286 \times \sqrt{V_{\alpha }}}{0.101 \times \sqrt{V_{P}}}$$0.101 \times \sqrt{V_{P}} =  0.286 \times \sqrt{V_{\alpha }}$$(0.101)^{2} \times V_{P} = (0.286)^{2}V _{\alpha} $$\dfrac{(0.101)^{2}\nu }{(0.286)^{2}} = V _{\alpha }$                 $(\because V_{P}=V)$$\Rightarrow  V_{\alpha } = \dfrac{V}{8}  volt$So, the answer is option (A).

Question 3

The momentum of a proton is p. The corresponding wavelength is

Accepted Answer

$\lambda =\dfrac{h}{mv}$   $=\dfrac{h}{p}$

Question 4

Moving with the same velocity, one of the following has the longest de Broglie wavelength

Accepted Answer

De-broglie wavelength$\lambda =\dfrac{h}{mv}$$\lambda \propto \dfrac{1}{m}$$\beta$ particle has least mass.So $\beta$ particle has the longest wavelength.

Question 5

Einstein&#8217;s photoelectric equation states that$E_{k} = h v - W$, In this equation $E_k$ refers to :

Accepted Answer

$E_{k}=h\nu  -\phi $ (here $\phi=W$)here $E_{k}$ refers to the maximum kinetic energy of emitted electrons.

Question 6

The wavelength of matter waves does not depend on

Accepted Answer

Wavelength of matter,$\lambda =\dfrac{h}{mv}$

Question 7

De-Broglie wavelength depends on

Accepted Answer

$\lambda =\dfrac{h}{mv}$So $\lambda $ depends on mass of the particle.

Question 8

If Planck&#8217;s constant is denoted by h and electronic charge by e, then photoelectric effect allows determination of:

Accepted Answer

In photoelectric effect we can know only the stopping potential.Stopping potential $=(\dfrac{hc}{\lambda }-\phi )/e$                                $=\dfrac{hc}{e\lambda }-\phi /e$So, we can determine only $h/e$ because we know stopping potential, frequency and work function of the metal.

Question 9

Matter waves are:

Accepted Answer

Correct Answer: option (D)Matter waves are waves associated with every moving particle and therefore are neither mechanical nor electromagnetic waves.These waves have $\lambda = h\times p$Where $\lambda$ is the wavelength, $h$ is plank's constant, and $p$ is the particle&#8217;s moment. Therefore, matter waves are neither mechanical nor electromagnetic waves.

Question 10

The Einstein&#8217;s photo-electric equation is based upon the conservation of

Accepted Answer

Einsteins photoelectric equations states that$E_{k}=h\nu - \phi $where, $E_{k}$ is maximum kinetic energy of emitted photoelectronsh is plank's constant.$\nu $ is frequency of incident light$\phi $ is work function of metal.So, it is clearly based upon energy conservation.

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An electron beam after collision with the target produces X-rays of wavelength $4 A^{0}$ . The velocity of the electron beam is

A proton when accelerated through a potential difference of V volt has a wavelength $λ$ associated with it. An $α$ - particle in order to have the same wavelength $λ$ must be accelerated through a p.d. of

The momentum of a proton is p. The corresponding wavelength is

Moving with the same velocity, one of the following has the longest de Broglie wavelength

Einstein’s photoelectric equation states that
$E_{k} = h v - W$ , In this equation $E_{k}$ refers to :

The wavelength of matter waves does not depend on

De-Broglie wavelength depends on

If Planck’s constant is denoted by h and electronic charge by e, then photoelectric effect allows determination of:

Matter waves are:

The Einstein’s photo-electric equation is based upon the conservation of

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An electron beam after collision with the target produces X-rays of wavelength 4A0. The velocity of the electron beam is

A proton when accelerated through a potential difference of V volt has a wavelength λ associated with it. An α - particle in order to have the same wavelength λ must be accelerated through a p.d. of

The momentum of a proton is p. The corresponding wavelength is

Moving with the same velocity, one of the following has the longest de Broglie wavelength

Einstein’s photoelectric equation states that Ek​=hv−W, In this equation Ek​ refers to :

The wavelength of matter waves does not depend on

De-Broglie wavelength depends on

If Planck’s constant is denoted by h and electronic charge by e, then photoelectric effect allows determination of:

Matter waves are:

The Einstein’s photo-electric equation is based upon the conservation of

Physics
3647 Views

An electron beam after collision with the target produces X-rays of wavelength $4 A^{0}$ . The velocity of the electron beam is

A proton when accelerated through a potential difference of V volt has a wavelength $λ$ associated with it. An $α$ - particle in order to have the same wavelength $λ$ must be accelerated through a p.d. of

Einstein’s photoelectric equation states that
$E_{k} = h v - W$ , In this equation $E_{k}$ refers to :