On the QP and AM MC #37 shouldn't the correct answer be d) 4times wavelength

Rebecca,  lambda is equal to h/p. The energy of the electron for 600 V is U = KE = qV = 600q.

Use the relationship p^2/2m = KE. For the same electron, momentum, p, is proportional to the square root of KE.

For V = 150 V,  U = KE = qV = 150q, or 1/4th the initial KE

Since momentum is proportional to the square root of the KE, the new momentum is proportional to the square root of 1/4, which is 1/2.  Plug 1/2p in the h/p equation and we get a wavelength that is twice the original wavelength.

What to you think?

John

Rebecca,  lambda is equal to h/p. The energy of the electron for 600 V is U = KE = qV = 600q.

Use the relationship p^2/2m = KE. For the same electron, momentum, p, is proportional to the square root of KE.

For V = 150 V,  U = KE = qV = 150q, or 1/4th the initial KE

Since momentum is proportional to the square root of the KE, the new momentum is proportional to the square root of 1/4, which is 1/2.  Plug 1/2p in the h/p equation and we get a wavelength that is twice the original wavelength.

What to you think?

John