A metal whose threshold frequency is 9.02×1014 s-1 emits an electron with a velocity of 7.11×E5 m/s when radiation of 1.26×1015 s-1 strikes the metal.
Use these data to calculate the mass of the electron
To calculate the mass of the electron, we can use the equation for the kinetic energy of an electron emitted due to the photoelectric effect:
E = mv²/2
Where:
- E is the kinetic energy of the electron
- m is the mass of the electron
- v is the velocity of the electron
First, we need to find the kinetic energy of the electron using the given velocity. Plugging in the values, we get:
E = (7.11×10⁵ m/s)² / 2
Next, we need to find the energy of the incident radiation using the formula:
E = hf
Where:
- E is the energy of the radiation
- h is Planck's constant (6.62607015 × 10⁻³⁴ J·s)
- f is the frequency of the radiation
Plugging in the values, we have:
E = (6.62607015 × 10⁻³⁴ J·s) × (1.26×10¹⁵ s⁻¹)
Now, since the minimum energy required to emit an electron is given by the equation:
E = hf₀
Where:
- E is the energy of the incident radiation
- f₀ is the threshold frequency of the metal
We can rearrange this equation to solve for the threshold frequency:
f₀ = E / h
Plugging in the obtained value for E, we have:
f₀ = [(6.62607015 × 10⁻³⁴ J·s) × (1.26×10¹⁵ s⁻¹)] / (6.62607015 × 10⁻³⁴ J·s)
Now, we can use the equation for the energy of the electron emitted:
E = hf - hf₀
Rearranging it, we can solve for the mass of the electron:
m = 2E / v²
Plugging the values of E (the kinetic energy of the electron) and v (the velocity of the electron), we can calculate the mass of the electron.