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When photon of energy 4.0eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_A\: eV and de-Broglie wavelength \lambda _A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50eV is T_B = (T_A-1.5)eV. If the de-Broglie wavelength of these photoelectrons \lambda _B=2\lambda _A, then the work function(in eV) of metal B is: 
Option: 1 4
Option: 2 1.5
Option: 3 3
Option: 4 2

As we know \lambda =\frac{h}{\sqrt{2mT}}

where T=kinetic energy

So \frac{\lambda_A}{\lambda_B} = \sqrt{\frac{T_A}{T_B}}=\frac{1}{2}...(1)

Also Given \ \ T_B=T_A-1.5... (2)

Using equation (1) and (2)

T_A=2 \ eV , \ and \ T_B=1.5 \ eV

And using the Einstein equation

E_B=T_B+W_B\\ \Rightarrow W_B=E_B-T_B=(4.5-0.5)eV=4 \ eV

So the correct option is 1.

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Posted by

vishal kumar

A beam of electromagnetic radiation of intensity 6.4\times10^{-5} W/cm^2 is comprised of wavelength \lambda=310 nm. It falls normally on a metal (work function \phi=2 eV) of surface area 1 cm2. If one in 103 photons ejects an electron, total number of electrons ejected in 1s is 10x (hc=1240 eVnm, 1 eV=1.6 x 10-19J), then x is :-  
Option: 1 11
Option: 2 10
Option: 3 12
Option: 4 13

Energy in 1 second = 6.4 \times 10 ^{-5} \times \1 \ \times 1 = 6.4 \times 10 ^{-5} Joule

Energy (eV) = \frac{6.4 \times 10 ^{-5}}{1.6 \times 10^{-19}} = 4 \times 10^{14} eV

Energy \ of\ one\ photon = \frac{12400}{3100} = 4 \ eV

Number \ of \ photon \ = \frac{4 \times 10^{14}}{4} = 10^{14}

Number \ of \ electrons \ = \frac{10^{14}}{10^3} = 10^{11}

So, \ x = 11

So the correct answer is option 1.

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Posted by

Ritika Jonwal

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An electron of mass m and magnitude of charge \left | e \right | initially at rest gets accelerated by a constant electric field E.The rate of change of de-broglie wavelength of this electron at time t is ? ( ignoring relativistic effects is) :
Option: 1 \frac{-h}{\left | e \right |Et^{2}}      
Option: 2  \frac{\left | e \right |Et}{h}    
Option: 3  -\frac{h}{\left | e \right |Et}
Option: 4  -\frac{h}{\left | e \right |E\sqrt{t}}
 

 

 

\lambda _{D}=\frac{h}{mv}

\therefore\; \; v=at

v=\frac{eE}{m}t\; \; \; \; \left ( a=\frac{eE}{m} \right )

\lambda _{D}=\frac{h}{m\left ( \frac{eE}{m}t \right )}

\lambda _{D}=\frac{h}{eEt}

\frac{d\lambda _{d}}{dt}=-\frac{h}{\left | e \right |Et^{2}}

 

Hence the correct option is (1). 

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Posted by

avinash.dongre

An electron (mass m ) with initial velocity \vec{v} = v_{0}\widehat{i} + v_{0}\widehat{j} is in an electric field \vec{E} = -E_{0}\widehat{k}. If \lambda_{0} is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by :
Option: 1 \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}

Option: 2 \frac{\lambda_{0}}{\sqrt{2 + \frac{e^{2} E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}

Option: 3 \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2}t^{2}}{2m^{2}v_{0}^{2}}}}

Option: 4 \frac{\lambda_{0} \sqrt{2}}{\sqrt{1 + \frac{e^{2} E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}
 

 

 

  

\text { Initially } m\left(\sqrt{2} v_{0}\right)=\frac{h}{\lambda_{0}}

 

We know v=u+at and here a=\frac{qE}{m}=\frac{eE_0}{m}

 

\text { Velocity as a function of time }=v_{0} \hat{i}+v_{0} \hat{j}+\frac{e E_{0}}{m} t \hat{k}

 

So, magnitude of velocity:- \sqrt{2 v_{0}^{2}+\frac{e^{2} E_{0}^{2}}{m^{2}} t^{2}}$

 

\text { so wavelength } \lambda=\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{v}_{0}^{2}+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2}}{\mathrm{m}^{2}} \mathrm{t}^{2}}}

 

\Rightarrow \lambda=\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2}}{2\mathrm{m}^{2} \mathrm{v}_{0}^{2}} \mathrm{t}^{2}}}$

Hence the correct option is (3).

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Posted by

vishal kumar

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Radiation with wavelength 6561 \overset{o}{A} falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3\times 10^{-4}T. If the radius of the largest circular path followed by the electrons is 10mm, the work function (in eV)  of the metal is close to:
Option: 1 1.1
Option: 2 0.8
Option: 3 1.8
Option: 4 1.6
 

 

 

From photoelectric equation:- 

\\KE_{max} = E - \phi \\ = \frac{12400}{\lambda (\ln A)} - \phi \quad (in \ eV) \\\therefore r = \frac{\sqrt{2mKE}}{eB}

\\KE_{max} = \frac{r^2e^2B^2}{2m}

                   = \frac{r^2e2B^2}{2m} \quad (in \ eV)

\therefore \phi = \frac{12400}{6556} - \frac{r^2eB^2}{2m}

          =1.1 eV

Hence the option correct option is (1) .  

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Posted by

avinash.dongre

A particle moving with kinetic energy E has de Broglie wavelength \lambda. If energy \Delta E is added to its energy, the wavelength become \frac{\lambda}{2}. Value of \Delta E, is:
Option: 1 2E
Option: 2 4E
Option: 3 3E
Option: 4 E

 

 

 

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \\\Rightarrow \frac{\lambda_2}{\lambda_1} = \frac{\frac{\lambda}{2}}{\lambda} = \frac{1}{2} \\\Rightarrow \frac{E_1}{E_2} = \frac{E}{E + \Delta E} \\\Rightarrow \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{E_1}{E_2}} \\\Rightarrow \frac{1}{4} = \frac{E}{E + \Delta E} \\\Rightarrow \Delta E = 3E

Hence the option correct option is (3).  

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Posted by

avinash.dongre

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An electron (of mass m) & a photon have the same energy E. Find the ratio of de Broglie wavelength of an electron and the wavelength of the photon (c = speed of light in vacuum).  
Option: 1  \left ( \frac{E}{2m} \right )^{1/2} 
Option: 2   \frac{1}{c}\left ( \frac{2E}{m} \right )^{1/2}

Option: 3  c\left ( 2mE \right )^{1/2} 
Option: 4  \frac{1}{c}\left ( \frac{E}{2m} \right )^{1/2}
 

 

 

 

\lambda_{\mathrm{d}}$ for electron $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \\ \lambda$ for photon $=\frac{\mathrm{hC}}{\mathrm{E}}$ \\ Ratio $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\times \frac{\mathrm{E}}{\mathrm{hC}}=\frac{1}{\mathrm{C}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{m}}}

Hence the correct option is (4).  

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Posted by

Ritika Jonwal

A particle of mass 9.1\times 10^{-31} travels in a medium with a speed of 10^{6} m/s and a photon of a radiation of linear momentum 10^{-27\: }kg \, m/s travels in vacuum. The wavelength of photon is _______ times the wavelength of the particle.
 

\begin{aligned} &m=9.1 \times 10^{-31} \mathrm{~kg} \\ &V=10^{6} \mathrm{~m} / \mathrm{s} \end{aligned}

\begin{gathered} \text { Momentum of photon }=P=10^{-27} \mathrm{~kg}\left(\frac{\mathrm{m}}{\mathrm{s}}\right) \\ \qquad \begin{array}{c} P=\frac{h}{\lambda}=10^{-27} \\ \lambda_{\text {photon }}=\frac{h}{10^{-27}} \rightarrow(1) \end{array} \end{gathered}

\lambda_{\text {clectron }}=\frac{h}{mv}=\frac{h}{9.1 \times 10^{-25}} \rightarrow(2)

\begin{aligned} &\frac{\lambda_{\text {photon }}}{\lambda_{\text {electron }}}=\frac{h / 10^{-27}}{h / 9.1 \times 10^{-25}} \\ &\frac{\lambda_{\text {photon }}}{\lambda_{\text {electron }}}=\frac{9.1 \times 10^{-25}}{10^{-27}}=910 \end{aligned}

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vishal kumar

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In a photoelectric experiment, ultraviolet light of wavelength 280 nm is used with lithium cathode having work function \phi = 2\cdot 5eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. \left ( h= 6\cdot 63\times 10^{-34}Js,c= 3\times 10^{8}ms^{-1} \right )
Option: 1 1\cdot 1V
Option: 2 0\cdot 6V
Option: 3 1\cdot 3V
Option: 4 1\cdot 9V

\lambda_{1}= 280\, nm
\phi_{0}= 2\cdot 5\, ev
\lambda_{2}= 400\, nm
By Einstein's photoelectric eqn,
\frac{hc}{\lambda_{1}}= \phi_{0}+e\left ( v_{0} \right )_{1}
E\left ( eV \right )= \frac{1240}{\lambda\left ( nm \right )}
\left ( \frac{hc}{\lambda_{1}} \right )\left ( eV \right )= \frac{1240}{280}= \frac{31}{7}
\left ( \frac{hc}{\lambda_{1}} \right )\left ( eV \right )= 4\cdot 428\, eV
\left ( \frac{hc}{\lambda_{2}} \right )eV= \frac{1240}{400}= 3\cdot 1\, eV
\frac{hc}{\lambda_{1}}-\frac{hc}{\lambda_{2}}= \left ( \phi_{0}-\phi_{0} \right )+e\left ( v_{0} \right )_{1}-e\left ( v_{0} \right )_{2}
1\cdot 328\, eV= e\left [\left ( v_{0} \right )_{1} -\left ( v_{0} \right )_{2} \right ]
\Delta v_{0}= 1\cdot 328\, V
The correct option is (3)
 

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Posted by

vishal kumar

The de-Broglie wavelength of a paraticle having kinetic energy E is \lambda. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the intial value?
Option: 1 E
Option: 2 \frac{7}{9}E
Option: 3 \frac{16}{9}E
Option: 4 \frac{1}{9}E

De-Broglie\: wavelength\: \left ( \lambda \right )= \frac{h}{\sqrt{2mE}}
\lambda_{i}= \frac{h}{\sqrt{2mE}}\rightarrow \left ( 1 \right )
\lambda_{f}=0\cdot 75\lambda_{i}= \frac{h}{\sqrt{2mE_{f}}}\rightarrow \left ( 2 \right )
\frac{1}{0\cdot 75}= \frac{\sqrt{E_{f}}}{\sqrt{E}}
\left ( \frac{4}{3} \right )^{2}E= E_{f}
E_{f}= \frac{16}{9}E
Extra energy is given = E_{f}= E_{i}
                              = \frac{16}{9}E-E
                              = \frac{7E}{9}
The correct option is (2)  

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vishal kumar

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