Q: A photosensitive metallic surface has work function, h _{0} . If photons of energy 2h _{0} fall on this surface, the electrons come out with a maximum velocity of 4 10 ^{6} m/s. When the photon energy is increased to 5h _{0} , then maximum velocity of photo electrons will be

2hv_{0}=hv_{0}+\frac{1}{2}m(4\times 10^{6})^{2} 5hv_{0}=hv_{0}+\frac{1}{2}mv^{2}\Rightarrow 4 \times \frac{1}{2}m(4\times 10^{6}) =\frac{1}{2}mv^{2}\Rightarrow v = 8 \times 10^{6} m/s

Question 1

A particle having a de Broglie wavelength of 1.0  A^{0}  is associated with a momentum of (given  h = 6.6 	imes 10^{-34}  Js)

Accepted Answer

Momentum of a particle with given de Broglie wavalength             =\dfrac{h}{\lambda }             =\dfrac{6.6	imes 10^{-34}}{1	imes 10^{-10}}             ={6.6	imes 10^{-24}}\ kg \ m/s So, the answer is option (C).

Question 2

When a metallic surface is illuminate with radiation of wavelength  \lambda , the stopping potential is V. If the same surface is illuminated with radiation of wavelength  2\lambda , the stopping potential is  \displaystyle\frac{V}{4} . The threshold wavelength for the metallic surface is:

Accepted Answer

We know that: eV=\dfrac{hc}{\lambda}-\phi  ........ (1) \dfrac{eV}{4}=\dfrac{hc}{2\lambda}-\phi  ........ (2)From equations (1) and (2), we get: \Rightarrow 4=\dfrac{\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}}{\dfrac{1}{2\lambda}-\dfrac{1}{\lambda_0}} \Rightarrow \lambda_0=3\lambda

Question 3

Photons with energy  5 \ eV  are incident on a cathode  C  in a photoelectric cell. The maximum energy of emitted photo electrons is  2 \ eV . When photons of energy  6 \ eV  are incident. on  C , no photoelectrons will reach the anode  A , if the stopping potential of  A  relative to  C  is :

Accepted Answer

Case 1 :   Energy of photon     E = 5  eVMaximum kinetic energy of photoelectrons        K = 2  eVUsing      K = E -\phi                     where  \phi  is the work function 	herefore     2 = 5 -\phi                  \implies \phi = 3  eVCase 2 :     E = 6  eV 	herefore  Maximum kinetic energy of photoelectrons      K = 6 - 3 = 3   eVthus stopping potential of A relative to C is   - 3  V.

Question 4

When a metallic surface is illuminate with radiation of wavelength  \lambda , the stopping potential is V. If the same surface is illuminated with radiation of wavelength  2\lambda , the stopping potential is  \displaystyle\frac{V}{4} . The threshold wavelength for the metallic surface is:

Accepted Answer

We know that: eV=\dfrac{hc}{\lambda}-\phi  ........ (1) \dfrac{eV}{4}=\dfrac{hc}{2\lambda}-\phi  ........ (2)From equations (1) and (2), we get: \Rightarrow 4=\dfrac{\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0}}{\dfrac{1}{2\lambda}-\dfrac{1}{\lambda_0}} \Rightarrow \lambda_0=3\lambda

Question 5

The work function of the surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the :

Accepted Answer

E_{incident} = W + K_{max} ; K_{max} = eV_{0} hv = hv_{0} + eV_{0} ; stopping  potential = V_{0} \dfrac{hc}{\lambda }=6.2e + 5e  \lambda =\dfrac{h	imes c}{11.2 	imes 1.6 	imes 10^{-19}}=\dfrac{6.63 	imes 10^{-34}	imes 3	imes 10^{8}}{11.2 	imes 1.6 	imes 10^{-19}} = 1.1 	imes 10^{-7} m Hence the wavelength of ultraviolet region.

Question 6

A photosensitive metallic surface has work function, h  _{0} . If photons of energy 2h  _{0}  fall on this surface, the electrons come out with a maximum velocity of 4  10 ^{6}  m/s. When the photon energy is increased to 5h _{0} , then maximum velocity of photo electrons will be

Accepted Answer

2hv_{0}=hv_{0}+\frac{1}{2}m(4	imes 10^{6})^{2} 5hv_{0}=hv_{0}+\frac{1}{2}mv^{2}\Rightarrow 4 	imes \frac{1}{2}m(4	imes 10^{6}) =\frac{1}{2}mv^{2}\Rightarrow v = 8 	imes 10^{6} m/s

Question 7

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) :  (m_e = 9 	imes 10^{-31} kg)

Accepted Answer

For an electron accelerated through a potential V, \lambda = \dfrac{12.27}{\sqrt{V}}  \mathring { A }  = \dfrac{12.27 	imes 10^{-10}}{\sqrt{10000}}  = 12.27 	imes 10^{-12} m

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frequency vs potential

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De-broglie wavelength