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Sagot :
broglie wavelength of this electron is 107.5pm.
what is de broglie wavelength?
The wavelength (λ) that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength.
Given:
de broglie wavelength of this electron? [ke[ke = 12mv2;12mv2; 1 electron volt (ev)(ev) = 1. 602×10−19j] is 107.5pm
[tex]h = 6.63 \times 10 { }^{ - 34} [/tex]
[tex]k = 120ev[/tex]
[tex]me = 9 \times 10 {}^{ - 31} kg[/tex]
[tex]1ev = 1.6 \times 10 {}^{ - 19} [/tex]
[tex]p = \sqrt{2mk} [/tex]
[tex] = \sqrt{2 \times 9 \times 10 { }^{ - 31} \times 1.6 \times 10 {}^{ - 19} } [/tex]
[tex] = 5.88 \times 10 {}^{ - 24} kgm/s[/tex]
de broglie wavelength
λ=h/p
[tex] = \frac{6.63 \times 10 {}^{ - 34} }{5.88 \times 10 {}^{ - 24} } [/tex]
Thus from the above calculation the de broglie wavelength comes out to be 107.5pm
learn more about de broglie wavelength from here: https://brainly.com/question/28165547
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