In a TV tube an electric potential difference accelerates electrons from a rest position towards a screen. Just before reaching the screen, the electrons have a wavelength of 1.2×10−11?. Determine the kinetic energy of the electrons just before they reach the screen. Your final answer should have units of eV. Assume: ℎ=6.63×10−34?•? ?=3.00×108?/? ??=9.11×10−31?
In a TV tube an electric potential difference accelerates electrons from a rest position towards a screen. Just before reaching the screen, the electrons have a wavelength of 1.2×10−11?. Determine the kinetic energy of the electrons just before they reach the screen. Your final answer should have units of eV. Assume: ℎ=6.63×10−34?•? ?=3.00×108?/? ??=9.11×10−31?
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In a TV tube an electric potential difference accelerates electrons from a rest position towards a screen. Just before reaching the screen, the electrons have a wavelength of 1.2×10−11?. Determine the kinetic energy of the electrons just before they reach the screen. Your final answer should have units of eV.
Assume:
ℎ=6.63×10−34?•?
?=3.00×108?/?
??=9.11×10−31??
I attached my answer but am unsure of whether it is right.
Transcribed Image Text:1 = 1.2 × 10 -11m
h = 6.63 × 10 -34 j . s
m = 9.11 × 10 -3kg
v = ?
mv
h
V =
mh.
6.63×10 ~34 J.s
(9.11x10
*kg)(1.2×10 ,
'm)
= 6.06 × 10 7m/s
v = 6.06 × 10 7m/s
m = 9.11 × 10 -31kg
E = ?
E = {mv ²
-3
= }(9.11 × 10
(3.35 x 10 -15)
= 1.67 × 10 ¯15 J
'kg)(6.06 × 10 ’m/s) ²
= 1.67 ×10 ~15 ,
1.60 × 10 -19ev
= 1.05 × 10 “eV
Therefore, the kinetic energy of the electrons just before they reach the screen is 1.05 × 10 “eV.
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