An electron has de Broglie wavelength 2.83×10 10 m . Part A Determine the magnit

Question

An electron has de Broglie wavelength 2.83×1010 m .

Part A

Determine the magnitude of the electron's momentum pe.

Express your answer in kilogram meters per second to three significant figures.

Hints

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Part B

Determine the kinetic energy Ke of the electron.

Express your answer in joules to three significant figures.

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Significant Figures Feedback: Your answer 3.061018 = 3.06×1018 J was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your answer.

Part C

Determine the electron's kinetic energy in electron volts.

Express your answer in electron volts to three significant figures.

Hints

Hint 1. The relation between electron volts and joules(click to open)

An electron has de Broglie wavelength 2.83×1010 m .

Part A

Determine the magnitude of the electron's momentum pe.

Express your answer in kilogram meters per second to three significant figures.

Hints

pe = 2.34×1024   kgm/s  

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Correct

Part B

Determine the kinetic energy Ke of the electron.

Express your answer in joules to three significant figures.

Hints

Ke = 3.01×1018   J  

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Correct

Significant Figures Feedback: Your answer 3.061018 = 3.06×1018 J was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your answer.

Part C

Determine the electron's kinetic energy in electron volts.

Express your answer in electron volts to three significant figures.

Hints

Hint 1. The relation between electron volts and joules(click to open)

Ke =   eV  

Explanation / Answer

Part A -

Wavelength= 2.83 x 10-10 m
m= 9.1093821 x 10^(-31) kg
h = 6.626*10^(-34) J/s

As per de Broglie's expression -  
wavelength = (h/mv)
velocity = 6.626*10^(-34)/(9.1094*10^(-31) * 2.83 *10^(-10) )
= 0.257 x 10^7 m/s
momentum = mass * velocity
Pe = 9.1094 x 10^-31 x 0.257 x 10^7 = 2.34 x 10^-24 kgm/s

Part B

kinetic energy = 1/2 mv^2
= 0.5 *9.1094*10^(-31)*(0.257 x 10^7)^2
= 0.3008 x 10^-17 = 3.008 x 10^-18 J

Part C

Kinetic energy in eV = (3.008 x 10^-18) x 6.2415 x 10^18 = 18.77 eV.

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