Compton displacement in compton experiment of X-rays at 60o scattering angle is 0.0121 bohr radius. Calculate the planck constant.
The Compton displacement (\( \Delta \lambda \)) is the change in wavelength of the X-rays before and after scattering. It is related to the scattering angle (\( \theta \)) and the Compton wavelength of the electron (\( \lambda_c \)) by the formula:
\[ \Delta \lambda = \lambda_c \cdot (1 - \cos(\theta)) \]
where:
\( \Delta \lambda \) = Compton displacement,
\( \theta \) = scattering angle,
\( \lambda_c \) = Compton wavelength of the electron.
We are given that the Compton displacement (\( \Delta \lambda \)) is 0.0121 Bohr radius. The Compton wavelength of the electron is approximately \( 0.0282 \) Ångström (corresponding to 0.0282 x \( 10^{-10} \) meters).
Now, let's calculate the scattering angle (\( \theta \)) using the formula:
\[ \theta = \cos^{-1} \left(1 - \frac{\Delta \lambda}{\lambda_c}\right) \]
Substitute the given values:
\[ \theta = \cos^{-1} \left(1 - \frac{0.0121 \text{ Bohr radius}}{0.0282 \text{ Ångström}}\right) \]
\[ \theta \approx \cos^{-1}(0.572695) \approx 54.167^\circ \]
Now, to calculate the Planck constant (\( h \)), we can use the relationship between the Compton wavelength (\( \lambda_c \)) and the Planck constant:
\[ \lambda_c = \frac{h}{m_e \cdot c} \]
where:
\( m_e \) = mass of the electron,
\( c \) = speed of light.
The mass of the electron (\( m_e \)) is approximately \( 9.10938356 \times 10^{-31} \) kilograms.
Let's rearrange the equation to solve for \( h \):
\[ h = \lambda_c \cdot m_e \cdot c \]
Substitute the values:
\[ h = 0.0282 \times 10^{-10} \text{ meters} \times 9.10938356 \times 10^{-31} \text{ kg} \times 3.00 \times 10^8 \text{ m/s} \]
\[ h \approx 8.052 \times 10^{-35} \text{ J·s} \]
So, the Planck constant is approximately \( 8.052 \times 10^{-35} \) J·s.
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