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Quantum Physics and Atomic Models Unit

PROGRESSIVE SCIENCE INITIATIVE® (PSI®)

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3 Comments

Rebecca ODette • 3 years, 1 month agologin to reply

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

Melissa Axelsson • 3 years, 1 month agologin to reply

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

John Ennis • 3 years, 1 month agologin to reply

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

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