Question

Calculate the kinetic energy of the ejected electron when ultra-violet radiation of frequency 1.6 × 10^15/s strikes the surface of potassium metal. Threshold frequency of potassium is 5×10^14/s (h=6.63×10^-34Js)

Answer 1

Answer:

K.E = 72.93× 10^-20

Explanation:

we know,

hf=hf°+K.E

h(f-f°)=K.E

6.63×10^-34(1.6×10^15-5×10^14)=K.E

6.63×10^-34×11×10^14=K.E

72.93×10^-20=K.E