Answer to Question #81824 in Physics for physics question

Question

1. Protons in an accelerator at the Fermi National Laboratory near Chicago are accelerated to a total energy that is 400 times their rest energy.

(a) What is the speed of these protons? (Round your answers to six decimal places.)
E = 400 × mc2 = γmc2

(b) What is their kinetic energy?
( m p = 938.3 MeV/c 2 )

2. When light of wavelength 210nm falls on a gold surface, electrons having a maximum kinetic energy of 0.81 eV are emitted. Find values for the following.

(a) the work function of gold

(b) the cutoff wavelength

(c) the frequency corresponding to the cutoff wavelength

Answer 1

a)

400=1/√(1-v^2/c^2 )

v=0.999997c.

b)

K=(400-1)(938.3)=3744000 MeV=374.4 GeV.

2. When light of wavelength210nm falls on a gold surface, electrons having a maximum kinetic energy of 0.81 eV are emitted. Find values for the following.

(a)the work function of gold

(b)the cutoff wavelength

Solution

a) Energy of a single photon in light of wavelength 220nm:

E= hf= h c/λ =1241.5/210 =5.91 eV

The work function of gold:

5.91-0.81=5.10 eV.

b)

λ_cutoff=1241.5/5.10 =243 nm