1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 5. A X-ray photon of $$ \left(\lambda=11 \mathrm{~A}^{\wedge} ight) $$ collides with a stationary free electron, the wavelength of the scattered $$ X $$-ray photon increases to $$ 19 A^{\wedge} $$. Then, the velocity of the electron becomes $$ \qquad $$ $$ 7.4 imes 10^{\wedge}(-3) \mathrm{m} / \mathrm{s} $$ $$ 1.29 imes 10^{\wedge}(2) \mathrm{m} / \mathrm{s} $$ $$ 1.29 imes 10^{\wedge}(7) \mathrm{m} / \mathrm{s} $$ $$ 7.41 imes 10^{\wedge}(2) \mathrm{m} / \mathrm{s} $$ NEXT QUESTION

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 5. A X-ray photon of $$ \left(\lambda=11 \mathrm{~A}^{\wedge}ight) $$ collides with a stationary free electron, the wavelength of the scattered $$ X $$-ray photon increases to $$ 19 A^{\wedge} $$. Then, the velocity of the electron becomes $$ \qquad $$ $$ 7.4 	imes 10^{\wedge}(-3) \mathrm{m} / \mathrm{s} $$ $$ 1.29 	imes 10^{\wedge}(2) \mathrm{m} / \mathrm{s} $$ $$ 1.29 	imes 10^{\wedge}(7) \mathrm{m} / \mathrm{s} $$ $$ 7.41 	imes 10^{\wedge}(2) \mathrm{m} / \mathrm{s} $$ NEXT QUESTION

UpStudy AI · Accepted Answer

We first note that when the $$X$$-ray photon collides with a free electron and its wavelength increases, the lost photon energy is transferred to the electron as kinetic energy. We can find the kinetic energy $$K$$ of the electron from the energy difference of the photon before and after scattering.

**Step 1. Write the photon energies**

The energy of a photon is given by
$$
E=\frac{hc}{\lambda}\,,
$$
where
- $$h$$ is Planck's constant,
- $$c$$ is the speed of light,
- $$\lambda$$ is the wavelength.

A commonly used value is $$hc\approx12400\,\text{eV}\cdot\text{\AA}$$. The initial and final energies are then
$$
E_i=\frac{12400}{11}\,\text{eV}\quad\text{and}\quad E_f=\frac{12400}{19}\,\text{eV}\,.
$$

**Step 2. Compute the energies**

Calculate the initial energy:
$$
E_i\approx\frac{12400}{11}\approx1127.27\,\text{eV}\,.
$$
Calculate the final energy:
$$
E_f\approx\frac{12400}{19}\approx652.63\,\text{eV}\,.
$$

**Step 3. Find the energy transferred to the electron**

The kinetic energy of the electron is the energy difference:
$$
K=E_i-E_f\approx1127.27-652.63\approx474.64\,\text{eV}\,.
$$

**Step 4. Convert the kinetic energy to Joules**

We use the conversion factor $$1\,\text{eV}=1.602\times10^{-19}\,\text{J}$$:
$$
K\approx474.64\,\text{eV}\times1.602\times10^{-19}\,\frac{\text{J}}{\text{eV}}\approx7.61\times10^{-17}\,\text{J}\,.
$$

**Step 5. Determine the electron’s velocity**

The kinetic energy is related to the velocity $$v$$ of the electron by
$$
K=\frac{1}{2}m_e v^2\,,
$$
where $$m_e\approx9.11\times10^{-31}\,\text{kg}$$ is the electron mass. Solving for $$v$$, we have
$$
v=\sqrt{\frac{2K}{m_e}}\,.
$$
Substitute the numerical values:
$$
v=\sqrt{\frac{2\times7.61\times10^{-17}\,\text{J}}{9.11\times10^{-31}\,\text{kg}}}\,.
$$

Perform the calculation:
$$
v=\sqrt{\frac{1.522\times10^{-16}}{9.11\times10^{-31}}}
=\sqrt{1.67\times10^{14}}
\approx1.29\times10^{7}\,\text{m/s}\,.
$$

Thus, the velocity of the electron after collision is
$$
\boxed{1.29\times10^{7}\,\text{m/s}}\,.
$$

This corresponds to the option $$1.29 \times 10^{\wedge}(7)\,\mathrm{m}/\mathrm{s}$$.
The electron's velocity after the collision is $$1.29 \times 10^{7}\,\mathrm{m/s}$$.

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5. A X-ray photon of collides with a stationary free electron, the wavelength of the scattered -ray photon increases to . Then, the velocity of the electron becomes

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 5. A X-ray photon of collides with a stationary free electron, the wavelength of the scattered -ray photon increases to . Then, the velocity of the electron becomes NEXT QUESTION