Question

assignment

• compare the wavelength for an electron (mass = 9.11 x 10-31 kg) travelling at a speed of 1.0 x 107 m/s with that for a ball (mass = 0.10 kg) travelling at 35 m/s.
• use $\lambda = \frac{h}{mv}$
• $h = 6.626 x 10-34 js (kgm2/s)$
• which wavelength is smaller?

Explanation:

Step1: Calculate wavelength of electron

Using the formula \(\lambda = \frac{h}{mv}\), where \(h = 6.626\times 10^{-34}\space Js\), \(m_{electron}=9.11\times 10^{-31}\space Kg\), \(v_{electron}=1.0\times 10^{7}\space m/s\).
\(\lambda_{electron}=\frac{6.626\times 10^{-34}}{(9.11\times 10^{-31})\times(1.0\times 10^{7})}\)
First, calculate the denominator: \((9.11\times 10^{-31})\times(1.0\times 10^{7}) = 9.11\times 10^{-24}\)
Then, \(\lambda_{electron}=\frac{6.626\times 10^{-34}}{9.11\times 10^{-24}}\approx 7.27\times 10^{-11}\space m\)

Step2: Calculate wavelength of ball

Using the same formula, where \(m_{ball}=0.10\space Kg\), \(v_{ball}=35\space m/s\).
\(\lambda_{ball}=\frac{6.626\times 10^{-34}}{(0.10)\times(35)}\)
Calculate the denominator: \(0.10\times35 = 3.5\)
Then, \(\lambda_{ball}=\frac{6.626\times 10^{-34}}{3.5}\approx 1.89\times 10^{-34}\space m\)

Step3: Compare the two wavelengths

Compare \(\lambda_{electron}\approx 7.27\times 10^{-11}\space m\) and \(\lambda_{ball}\approx 1.89\times 10^{-34}\space m\). Since \(1.89\times 10^{-34}<7.27\times 10^{-11}\), the wavelength of the ball is smaller.

Answer:

The wavelength of the ball (mass = 0.10 Kg travelling at 35 m/s) is smaller.