## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 28 - Problems - Page 1074: 6

#### Answer

$v = 1420~m/s$

#### Work Step by Step

We can find the required speed: $\lambda = \frac{h}{mv}$ $v = \frac{h}{m~\lambda}$ $v = \frac{6.626\times 10^{-34}~J~s}{(1.67\times 10^{-27}~kg)(0.28\times 10^{-9}~m)}$ $v = 1420~m/s$

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