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Engineering
101 Views   |

The wavelength $\lambda _{e}$ of an electron and $\lambda _{p}$ of a photon are of same energy E are related by

• Option 1)

$\lambda _{p}\alpha \lambda _{e}$

• Option 2)

$\lambda _{p}\alpha \sqrt{\lambda _{e}}$

• Option 3)

$\lambda _{p}\alpha \frac{1}{\sqrt{\lambda _{e}}}$

• Option 4)

$\lambda _{p}\alpha {\lambda _{e}}^{2}$

As we discussed in concept Energy of a photon - - wherein    & De - Broglie wavelength with charged particle -   - wherein    For photon For electron,     Option 1) This option is incorrect. Option 2) This option is incorrect. Option 3) This option is incorrect. Option 4) This option is correct.
Engineering
128 Views   |

The de-Broglie wavelength of neutron in thermal equilibrium at tempreture T is

• Option 1)

$\frac{30}{\sqrt{T}}A^{\circ}$

• Option 2)

$\frac{3.08}{\sqrt{T}}A^{\circ}$

• Option 3)

$\frac{0.308}{\sqrt{T}}A^{\circ}$

• Option 4)

$\frac{0.0308}{\sqrt{T}}A^{\circ}$

As we discussed in concept De - Broglie wavelength with charged particle -   - wherein       Option 1) This option is correct. Option 2) This option is incorrect. Option 3) This option is incorrect. Option 4) This option is incorrect.
Engineering
93 Views   |

A source of light is placed at a distance of 50 cm from a photocell and the stopping potential is found to be $V_{0}$. If the distance between the light source and photocell is made 25 cm, the new stopping potential will be

• Option 1)

$2V_{0}$

• Option 2)

$\frac{V_{0}}{2}$

• Option 3)

$V_{0}$

• Option 4)

$4V_{0}$

As we discussed in concept Stopping Potential /Cut-off Potential - It is defined as the potential necessary to stop any electron from reaching the other side. -    Stopping potential depends on frequency of light and not on intensity since frequency of light remains constant, stopping potential will remain same.   Option 1) This option is incorrect. Option 2) This option is...
Engineering
107 Views   |

For photoelectric emission from certain metal the cut-off frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)

• Option 1)

$\sqrt{\frac{hv}{m}}$

• Option 2)

$\sqrt{\frac{2hv}{m}}$

• Option 3)

$2\sqrt{\frac{hv}{m}}$

• Option 4)

$\sqrt{\frac{hv}{(2m)}}$

As we discussed in concept Conservation of energy -   - wherein     here         => Option 1) This option is incorrect. Option 2) This option is correct. Option 3) This option is incorrect. Option 4) This option is incorrect.
Engineering
916 Views   |

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:

• Option 1)

0.65 eV

• Option 2)

1.0 eV

• Option 3)

1.3 eV

• Option 4)

1.5 eV

Let Initial energy was E. Then abter increasing it by 20% it becomes 1.2 E . Let work function is  . Then   E-  = 0.5 ev ------------(1)          1.2 -  = 0.8 ev ------------(2) or     or  Option 1) 0.65 eV Incorrect Option 2) 1.0 eV Correct Option 3) 1.3 eV Incorrect Option 4) 1.5 eV Incorrect
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