The work function of a metal is 12 ev . If a radiation of wavelength 3000 A° is incident on it , the maximum K.E of emitted photo electrons is : -

The work function of a metal is 12 ev . If a radiation of wavelength 3000 A° is incident on it , the maximum K.E of emitted photo electrons is . . . . . . . . . . : -

A ) 2.3 x 10 ^ ( - 16 )  J  ,                          B ) 3.4 x 10 ^ ( - 19 )  J  , 

C ) 4.9 x 10 ^ ( - 23 )  J  ,                          D ) 9.5 x 10 ^ ( - 23 )  J  ,

Correct Answer : -   B ) 3.4 x 10 ^ ( - 19 )  J  , 

Solution : -

∴   h = 6.6 x 10 ^ ( - 34 )   J s  , 

∴   c = 3 x 10 ^ 8  m / s   , 

∴   1 ev = 1.6 x 10 ^ ( - 19 )  J .

So  , we have formula : -

E = Φ + K.Emax . . . . . . . . . . . . eq. 1 ,

So to find value of K.Emax  from above equation , first we have to find value E and Φ : -

E = hf = hc / λ = ( 6.6 x 10 ^ ( - 34 )  x  3 x 10 ^ 8 ) / ( 3000 x 10 ^ ( - 10 ) ) .

E =  19.8 x 10 ^ ( - 26 )  /  ( 3000 x 10 ^ ( - 10 ) ) .

E = 6.6 x 10 ^ ( - 19 )  J .

So , Now we will find Φ : -

Work function = Φ = hf˳ = 2 ev  =  2 x ( 1.6 x 10 ^ ( - 19 ) )  , 

Φ  =  3.2 x 10 ^ ( - 19 )  .

So  put value of Φ and E in eq. 1 : -

6.6 x 10 ^ ( - 19 ) = 3.2 x 10 ^ ( - 19 )  +  K.Emax  ,

K.Emax  =  6.6 x 10 ^ ( - 19 )  -  3.2 x 10 ^ ( - 19 )  ,

K.Emax  =  3.4 x 10 ^ ( - 19 )  J  ,  Answer .