Respuesta :
Answer: 1ev = 1.609*10^-19 J
velocity of electron = v = 2.97 * 10^7 m/s
Explanation: The magnitude of an electronic charge (e) is given as
e = 1.609*10^-19c
The amount of kinetic energy an electron has when it moves through 1v equals the work done used in accelerating the electron ( according to work- energy theorem).
K.E = W = eV
e = magnitude of electron = 1.609 * 10^-19c
V = 1v
Hence 1ev = 1.609 * 10^-19 * 1
1ev = 1.609 * 10^-19 J.
If K.E = 2500ev.
By converting this to joules we have that
If 1ev = 1.609 * 10^-19 J.
Then 2500ev = 1.609 * 10^-19 * 2500
K.E ( in joules) = 4005 * 10^-19
K.E = 4.005 * 10^-16 J.
But K.E = 1/2mv²
Where K.E = kinetic energy of electron = 4.005 * 10^-16 J.
m = mass of electronic charge = 9.11 * 10 ^-31kg.
By substituting the parameters, we have that
4.005 * 10^-16 = 1/2 * 9.11 * 10 ^-31 * v²
4.005 * 10^-16 * 2 = 9.11 * 10 ^-31 * v²
8.01 * 10^-16 = 9.11 * 10^-31 * v²
v² = 8.01 * 10^-16/ 9.11 * 10^-31
v² = 0.879 * 10^15
v² = 8.79 * 10^14
v = √8.79 * 10^14
v = 2.97 * 10^7 m/s
The value of 1eV in joules is 1.6 × 10⁻¹⁹ J
The required speed of proton is 2.97 × 10⁷ m/s
Calculating the speed of proton:
e = charge on electron = 1.6 × 10⁻¹⁹ C
Hence
1eV = 1.6 × 10⁻¹⁹ J
If according to the question, the kinetic energy of the proton is
K.E = 2500 eV.
Converting to joules we get :
2500eV = 1.6 × 10⁻¹⁹ × 2500
K.E = 4.005 × 10⁻¹⁶J.
Also, we know that
KE = ¹/₂mv²
here, m = mass of electron = 9.11 × 10⁻³¹ kg.
By substituting the parameters, we get:
[tex]4.005 \times 10^{-16} = 1/2 \times 9.11 \times 10 ^{-31}\times v^2\\\\8.01 \times 10^{-16} = 9.11 \times 10^{-31} \times v^2\\\\v^2 = 8.01 \times 10^{-16}/ 9.11 \times 10^{-31}\\\\v^2 = 0.879 \times 10^{15}\\\\v^2 = 8.79 \times 10^{14}\\\\[/tex]
v = 2.97 × 10⁷ m/s
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