Light of energy 1.12 x 10\(^{-18}\) J is incident on a metal of ejected electrons with maximum energy of 8.0 x 10\(^{-19}\)J. Calculate the threshold frequency of the metal. [h = 6.626 x 10\(^{-34}\)Js]
A
4.8 x 10\(^{14}\)Hz
B
1.2 x 10\(^{15}\)Hz
C
1.7 x 10\(^{15}\)Hz
D
2.8 x 10\(^{15}\)Hz
Verified Correct Answer: A.
Physical Principles & Numerical Solution
E = hf + W\(_o\)
E = 1.12 x 10\(^{-18}\)J
Max. Energy = 8.0 x 10\(^{-19}\)J.
W\(_o\) = E - Max. Energy = 1.12 x 10\(^{-18}\)J - 8.0 x 10\(^{-19}\)J. = 3.2 x 10\(^{-19}\)J.
But W\(_o\) = hf\(_o\)
f\(_o\) = \(\frac{W_o}{h}\) = \(\frac{3.2 \times 10^{-19}}{6.626 \times 10^{-34}}\) = 4.83 x 10\(^{14}\)Hz
Source: WAEC / JAMB / NECO Classroom Archive (Record #72058)