Question

The kinetic energy of the electron emitted by light of frequency 3.5 into 10 power 15 is your name metal x having threshold frequency 1.5 into 10 power 15

Answer 1

Answer:13.24×10^-17

Explanation:we know that

hv =hv° +K.E (v = frequency and v°=threshold frequency)

h =6.62×10^-32

6.62 ×10^-32 ×3.5 ×10^15 = 6.62 ×10^-32×1.5 ×10^15+K.E

6.62 ×10^-32×10^15 (3.5 -1.5 )= K.E

On calculating we get

13.24×10^-17

I think this is the right process

Answer 2

Answer:

$Frequency \: of \: light = 3.5 \times {10}^{15} \: Hz \\ threshold \: frequency = 1.5 \times {10}^{15} \: Hz \\ \\ kinetic \: energy \: = (energy \: of \: light) - (threshold \: energy) \\ = hv - hv_{0} = h(v - v_{0}) \\ = 6.6 \times {10}^{ - 34} \times ((3.5 \times {10}^{15} \: Hz) - (1.5 \times {10}^{15})) \\ = 6.6 \times {10}^{ - 34} \times ({10}^{15} \times (3.5 \: Hz - 1.5) \\ = 6.6 \times {10}^{ - 34} \times {10}^{15} \times 2 \\ = 13.2 \times {10}^{ - 34 + 15} \\ = 13.2 \times {10}^{ - 19}$