Question

The kinetic energy of photoelectron when a
radiation of 104 Hz frequency hit the metal is
(vo = 103 Hz)

(1) 2.14 x 10-18 J
(2) 6.73 x 10-24 J
43) 5.96 ~ 10-30 J
(4) 7.12 x 10-34 J​

Answer 1

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ANS=

Given that,

v

o

=7.0×10

14

s

−1

v=1.0×10

15

s

−1

As we know, hv=hv

0

+

2

1

mv

2

i.e. h(v−v

0

)=K.E.

∴K.E.=(6.626×10

−34

)×(1.0×10

15

−7.0×10

14

)=6.626×10

−34

×10

14

(10−7)

∴K.E=19.878×10

−20

J

Answer 2

Answer:

The concept used here is Einstein photo electric equation ie. E = ϕ + KE

Explanation:

As we know,

 E = ϕ + KE  and it can also be written as,

 hv = hv∘ +KE

or, h(v−v∘) = KE .........(1)

now we are given with v and v∘

v = 10^4 Hz

v∘ = 10^3 Hz

h= 6.62607015 × 10^-34 m2 kg / s  

So , put it in equation (1)

h(10^4 - 10^3)=KE              [ 10^4 - 10^3 = 9000 and its not 10^1 )

h ( 9000) = KE

KE = 5.96 * 10 ^-30 J

So final answe is 5.96 * 10 ^-30 J