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 .
