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 .
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