Question

Яку максимальну швидкість мають вирвані з літію електрони під час його опромінення світлом з довжиною хвилі 300 нм? Червона межа фотоефекту для літію 1,2мкм.​

Answer 1

Ответ:

см фото

Объяснение:

Answer 2

Maximum Speed of Ejected Electrons and Red Threshold for Photoelectric Effect in Lithium

The maximum speed of ejected electrons and the red threshold for the photoelectric effect in lithium can be calculated using the given information.

Maximum Speed of Ejected Electrons: The maximum speed of ejected electrons can be calculated using the formula:

\[ v_{\text{max}} = \sqrt{\frac{2E_{\text{max}}}{m}} \]

Where: - \( v_{\text{max}} \) = maximum speed of ejected electrons - \( E_{\text{max}} \) = maximum kinetic energy of the ejected electrons - \( m \) = mass of the electron

Red Threshold for Photoelectric Effect: The red threshold for the photoelectric effect can be calculated using the formula:

\[ \lambda_{\text{threshold}} = \frac{hc}{E_{\text{threshold}}} \]

Where: - \( \lambda_{\text{threshold}} \) = threshold wavelength - \( h \) = Planck's constant - \( c \) = speed of light - \( E_{\text{threshold}} \) = threshold energy for the photoelectric effect

Let's calculate the maximum speed of ejected electrons and the red threshold for the photoelectric effect in lithium using the given values.

Calculations:

Given: - Wavelength of light, \( \lambda = 300 \, \text{nm} = 300 \times 10^{-9} \, \text{m} \) - Red threshold for photoelectric effect for lithium, \( \lambda_{\text{threshold}} = 1.2 \, \mu\text{m} = 1.2 \times 10^{-6} \, \text{m} \)

Using the formula \( \lambda_{\text{threshold}} = \frac{hc}{E_{\text{threshold}}} \), we can solve for \( E_{\text{threshold}} \).

\[ E_{\text{threshold}} = \frac{hc}{\lambda_{\text{threshold}}} \]

Substitute the values: \[ E_{\text{threshold}} = \frac{6.626 \times 10^{-34} \, \text{J s} \times 3.00 \times 10^8 \, \text{m/s}}{1.2 \times 10^{-6} \, \text{m}} \]

\[ E_{\text{threshold}} \approx 4.965 \times 10^{-19} \, \text{J} \]

Now, using the formula \( v_{\text{max}} = \sqrt{\frac{2E_{\text{max}}}{m}} \), we can solve for \( v_{\text{max}} \).

Given: - Mass of the electron, \( m = 9.109 \times 10^{-31} \, \text{kg} \)

\[ v_{\text{max}} = \sqrt{\frac{2 \times 4.965 \times 10^{-19} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} \]

\[ v_{\text{max}} \approx 6.61 \times 10^5 \, \text{m/s} \]

Results:

- Maximum speed of ejected electrons: Approximately 6.61 x 10^5 m/s - Red threshold for photoelectric effect for lithium: Approximately 1.2 μm

Therefore, the maximum speed of ejected electrons from lithium under the given conditions is approximately 6.61 x 10^5 m/s, and the red threshold for the photoelectric effect for lithium is approximately 1.2 μm.