Question

Howard Wolowitz is developing a radiation detector for the Intemational space Station. He uses a thin film composed of a metal alloy that has a threshold frequency of 5.5x10^15 Hz. When radiation is shone on the thin film, the fastest electron emitted has a velocity of 668000 m/s.

A. What is the KE, in J, of the fastest ejected electron?

I was able to figure this one out which was 2.03x10^-19

B. What is the wavelength, in nm , of the photon that caused the ejection of this electron?

I tried using wavelength= h/mv. But I didn't get the right answer. The right answer for this is 52nm. Which I do not know how to get!

C. What is the wavelength, in nm, of this electron?

The right answer for this one is 1.09 nm

I just need guidance for answering these question. If you are able to help thank you!

Answer 1

c. lambdfa = h/mv = 1.09 nm
b. The threshold is 5.5E15 Hz
c = freq x wavelength gives me about 54 nm.

Answer 2

A. ok I got 2.09*10^-19 J but used crude electron mass approx

B. Ephoton = Eelectron + h * 5.5*10^15
Ephoton = 2.03*10^-19 + 6.63*10^-34*5.5*10^15
Ephoton = 2.03*10^-19 + 36.5*10^-19
Ephoton = 38.5 *10^-19 Joules
now what wavelength is that
E = h c/lambda so lambda = h c/E
lambda = 6.63*10^-34 * 3*10^8 / 38.5*10^-19
= .516*10^-7 = 51.6 *10^-9 meters
= 51.6 nanometers so I agree with them

Answer 3

for electron
lambda = h/mv = 6.63*10^-34 /[ 9.1*10^-31*6.68*10^5]
= .109*10^-8 = 1.09 * 10-9 = 1.09 nm