Question

An electron with a speed of 1.5×106m/s collides with an atom of element X. Shortly afterward, the atom emits a 1240 nm photon. What was the electron's speed after the collision? Assume that, because the atom is so much more massive than the electron, the recoil of the atom is negligible. Hint: The energy of the photon is not the energy transferred to the atom in the collision.

Answer 1

Answer:

The electron's speed after the collision is $1.61\times 10^6\ m/s$.

Explanation:

We have,

Initial speed of an electron, $u=1.5\times 10^6\ m/s$

Wavelength of a photon, $\lambda=1240\ nm$

Let v is the electron's speed after the collision. The total energy of the electron remains conserved in this case. The loss of energy of electron is equal energy of the emitted photon such that :

$\dfrac{1}{2}m(v^2-u^2)=\dfrac{hc}{\lambda}$

m is mass of electron

Putting all the values,

$\dfrac{1}{2}m(v^2-u^2)=\dfrac{hc}{\lambda}\\\\v^2=\dfrac{2hc}{m\lambda}+u^2\\\\v^2=\dfrac{2\times 6.63\times 10^{-34}\times 3\times 10^8}{9.1\times 10^{-31}\times 1240\times 10^{-9}}+(1.5\times 10^6)^2\\\\v=1.61\times 10^6\ m/s$

So, the electron's speed after the collision is $1.61\times 10^6\ m/s$.

Answer 2

The speed of electron after the collision is $1.61 \times 10^{6}\;\rm m/s$.

Given data:

The speed of electron before collision is, $u = 1.5 \times 10^{6} \;\rm m/s$.

The wavelength of photon is, $\lambda =1240 \;\rm nm = 1240 \times 10^{-9} \;\rm m$.

Let v is the electron's speed after the collision. The according to conservation of energy concept, "the total energy of the electron remains conserved". Hence, the loss of energy of electron is equal energy of the emitted photon such that :

Energy loss of electron = Energy emitted by photon

$\dfrac{1}{2}m(v^{2}-u^{2})=\dfrac{h \times c}{\lambda}$

Here, h is the Planck's constant, m is the mass of electron and c is the speed of light.

Solving as,

$\dfrac{1}{2} \times 9.1 \times 10^{-31} \times (v^{2}-(1.5 \times 10^{6})^{2})=\dfrac{6.63 \times 10^{-34} \times (3 \times 10^{8})}{1240 \times 10^{-9}}\\\\v=\sqrt{2.60 \times 10^{12} }\\\\v= 1.61 \times 10^{6}\;\rm m/s$

Thus, the speed of electron after the collision is $1.61 \times 10^{6}\;\rm m/s$.

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