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Engineering
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   If \lambda _{0} and \lambda be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

  • Option 1)

    \sqrt{\frac{2h}{m}\left ( \lambda _{0}-\lambda \right )}

  • Option 2)

    \sqrt{\frac{2hc}{m}\left ( \lambda _{0}-\lambda \right )}

  • Option 3)

    \sqrt{\frac{2hc}{m}\left ( \frac{\lambda _{0}-\lambda }{\lambda \lambda _{0}} \right )}

  • Option 4)

    \sqrt{\frac{2h}{m}\left ( \frac{1}{\lambda _{0}}-\frac{1}{\lambda } \right )}

 

As discussed in the concept

Photoelectric Effect -

\frac{1}{2}mu^{2}= hv-hv_{0}

- wherein

where

m is the mass of the electron

u is the velocity associated with the ejected electron.

h is plank’s constant.

v is frequency of photon,

v0 is threshold frequency of metal.

 

 v^{2} = \frac{2hc}{m}\left ( \frac{1}{\lambda } - \frac{1}{\lambda _{0}} \right )

v =\sqrt{ \frac{2hc}{m}\left ( \frac{1}{\lambda } - \frac{1}{\lambda _{0}} \right )}

 

Hence the correct option is 3

 


Option 1)

\sqrt{\frac{2h}{m}\left ( \lambda _{0}-\lambda \right )}

Option 2)

\sqrt{\frac{2hc}{m}\left ( \lambda _{0}-\lambda \right )}

Option 3)

\sqrt{\frac{2hc}{m}\left ( \frac{\lambda _{0}-\lambda }{\lambda \lambda _{0}} \right )}

Option 4)

\sqrt{\frac{2h}{m}\left ( \frac{1}{\lambda _{0}}-\frac{1}{\lambda } \right )}

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Engineering
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 Chloro compound of Vanadium has only spin magnetic moment of 1.73 BM.  This Vanadium chloride has the formula :

(at. no. of V=23)

  • Option 1)

    VCl2

  • Option 2)

    VCl4

  • Option 3)

    VCl3

  • Option 4)

    VCl5

 

As discussed in

Magnetic Quantum Number (m) -

It  gives information about the spatial orientation of the orbital with respect to standard set of co-ordinate axis.

-

 

 The value of magnetic moment is 1.73 BM

1.73 = \sqrt{n\left ( n+2 \right )} where n is the number of unpaired electrons

3 = n \left ( n+2 \right )

After calculation n = 1

V_{23} => 1S^{2},2S^{2},2p^{6},3S^{2},3p^{6},4S^{2},3d^{3}

To obtain one unpaired electron V should be tetrapositive ion and the formula of its chlorid should be VCl_{4}

Option 2 is correct

 

 

 

 


Option 1)

VCl2

Incorrect option 

Option 2)

VCl4

Correct option

Option 3)

VCl3

Incorrect option

Option 4)

VCl5

Incorrect option

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Engineering
91 Views   |  

The energy absorbed in the transition of the electron from n=2 to 1 in H-atom will be 

  • Option 1)

    10.2 meV

  • Option 2)

    3.4 meV

  • Option 3)

    13.6 meV

  • Option 4)

    12.75 meV

 
As we learnt in  Lyman Series spectrum -   Where This lies in Ultraviolet region -     where for Lyman series Z=1 (hydrogen) Rydberg constant, R = Energy absorbed = =        = 10.2 meV Option 1) 10.2 meV Option 2) 3.4 meV Option 3) 13.6 meV Option 4) 12.75 meV
Engineering
94 Views   |  

Element with electronic configuration 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}3p^{6}3d^{10},4s^{2}4p^{6}4d^{10}, 5s^{2}5p^{3} belongs to which group of the periodic table?

  • Option 1)

    3rd

  • Option 2)

    15th

  • Option 3)

    7th

  • Option 4)

    2nd

 
As we learnt in  Aufbau Principle - In the ground state of the atoms, the orbitals are filled in order of their increasing energies. - wherein    Given configuration: Element of the above configuration belongs in p block and for p block elements, group no = valence electrons...
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