## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 27 - Problems - Page 1041: 9

#### Answer

We can rank the situations in order of the stopping potential, from largest to smallest: $e \gt a = f \gt b = c \gt d$

#### Work Step by Step

Let $\lambda_0 = 50~nm$. We can write an expression for $K_{max}$, the maximum kinetic energy of an electron in the experiment: $K_{max} = \frac{hc}{\lambda_0}-\phi = \frac{hc}{50~nm}-\phi = E_0-\phi$ We can write an expression for $K_{max}$ in each case: (a) $K_{max} = \frac{hc}{200~nm}-\phi = \frac{1}{4}\times E_0-\phi$ (b) $K_{max} = \frac{hc}{250~nm}-\phi = \frac{1}{5}\times E_0-\phi$ (c) $K_{max} = \frac{hc}{250~nm}-\phi = \frac{1}{5}\times E_0-\phi$ (d) $K_{max} = \frac{hc}{300~nm}-\phi = \frac{1}{6}\times E_0-\phi$ (e) $K_{max} = \frac{hc}{100~nm}-\phi = \frac{1}{2}\times E_0-\phi$ (f) $K_{max} = \frac{hc}{200~nm}-\phi = \frac{1}{4}\times E_0-\phi$ Note that the stopping potential is proportional to the value of $K_{max}$ in each case. We can rank the situations in order of the stopping potential, from largest to smallest: $e \gt a = f \gt b = c \gt d$

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