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tigry1 [53]
3 years ago
13

1) Determine the magnitude of energy for each of the blanks on the diagram. Give the correct values for 1A, 1B, and 1C.

Physics
1 answer:
morpeh [17]3 years ago
4 0

Answer:

Explanation: y’all taking the same test as me hahahahah I got the answers but I can’t attach the picture here so hit me up on snap daniela_0789

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An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
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Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

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\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

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\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

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