Question

Please answer as fast as possible. Q 25

Answer 1

Answer:

Energy of Incident Light:

E = hclamda

E = 6.6×10^-343×108200×10^-9

E = 9.9×10^-20J

Work Function:

W = hclamda×o

W = 6.6×10^-343×108250×10^-9

W = 7.92×10^-20J

Energy of the emitted electrons:

K.E = E-W

K.E = 9.9×10^-20J - 7.92×10^-20J

K.E = 19.8×10^-20J