What is the cutoff frequency for a metal surface that has a work function of 6.22 eV?

1 ⤊

A) 2.01 X 10 ^15 Hz
b) 3.01 x 10^15 Hz
c) 1.50 x 10^15 Hz
d) 5.02 x 10^15 Hz

2007-04-18 20:46:41 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics

E=h*f, where f = frequency and h = Planck's constant = 4.135*10^-15 eV*s

Therefore f = E/h = 6.22/4.125*10^-15 = 1.5*10^15 Hz

2007-04-18 20:59:56 · answer #1 · answered by gp4rts 7 · 1⤊ 0⤋

What Is Cutoff Frequency

2016-11-15 03:04:09 · answer #2 · answered by ? 4 · 0⤊ 0⤋