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You decide to roll a 0.10 kg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelength. If you roll it at 1.4x1^-3 m/s, what is its wavelength? Express your answer to two significant figures and include the appropriate units. How will the answer from above compare with the de Broglie wavelength of the high-speed electron that strikes the back face of one of the early models of a TV screen at 1/10 the speed of light (2x10^-11 m) ?

User JFerro
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1 Answer

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ANSWER:

(a) 4.73*10^-30 m

(b) 2.37*10^-19 times smaller

Explanation:

Given:

Mass of ball = m = 0.10 kg

Speed of ball = v = 1.4x10^-3 m/s

(a)

Since, de Broglie wavelength is given by:


\lambda=(h)/(mv)

Where, h is the Plank's Constant ( h = 6.626x10^-34 kg m^2/s). Therefore, de Broglie wavelength of the ball will be:


\begin{gathered} \lambda=(6.626\cdot10^(-34))/(0.10\cdot1.4\cdot10^(-3)) \\ \lambda_{\text{ball}}=4.73\cdot10^(-30) \end{gathered}

(b)


\begin{gathered} \lambda_(electron)=2\cdot10^(-11) \\ \frac{\lambda_{\text{ball}}}{\lambda_(electron)}=(4.73\cdot10^(-30))/(2\cdot10^(-11)) \\ \frac{\lambda_{\text{ball}}}{\lambda_(electron)}=2.37\cdot10^(-19) \end{gathered}

It means that the wavelength of the ball is 2.37*10^-19 times smaller

User PaulDaviesC
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