Question

A beam of electrons is incident on a system of excited hydrogen atoms. What minimum speed must the electrons have to cause the emission of light due to the transition in the hydrogen atoms from \(n=2\) to \(n=1 ?\) (In the collision of an electron with a hydrogen atom, you may neglect the recoil energy of the hydrogen atom, because it has a mass much greater than that of the electron.)

Short answer

Answer: The minimum speed of the electrons required is approximately \(7.66 \times 10^5 \ m/s\).

Step-by-step solution

Calculate the energy difference between the energy levels \(n=2\) and \(n=1\) of the hydrogen atom

The energy of an electron in a hydrogen atom is given by the formula: $$E_n = -\frac{13.6 eV}{n^2}$$ where \(E_n\) is the energy of the electron in the nth energy level and n is the principal quantum number. The energy difference between the energy levels \(n=2\) and \(n=1\) can be calculated as: $$\Delta E = E_1 - E_2 = -\frac{13.6 eV}{1^2} -\left(-\frac{13.6 eV}{2^2}\right)$$ $$\Delta E = 13.6 eV \left(1 - \frac{1}{4}\right) = 10.2 eV$$

Find the minimum kinetic energy of the incident electrons

To cause the emission of light due to the transition of hydrogen atoms from \(n=2\) to \(n=1\), the minimum kinetic energy of the incident electrons must be equal to the energy difference between these energy levels. $$K_e = 10.2 eV$$

Convert the kinetic energy to Joules

To find the speed of the electrons, we will need the kinetic energy in Joules. We can convert the kinetic energy from electron-volt (eV) to Joules (J) using the conversion factor: $$1\ eV = 1.6 \times 10^{-19}\ J$$ So, the kinetic energy in Joules is: $$K_e = 10.2 eV \times \frac{1.6 \times 10^{-19} J}{1 eV} = 1.632 \times 10^{-18}\ J$$

Calculate the minimum speed of the electrons

The kinetic energy of an electron can be calculated using the formula: $$K_e = \frac{1}{2}mv^2$$ where \(m\) is the mass of the electron (\(9.11 \times 10^{-31}\ kg\)) and \(v\) is its speed. To find the minimum speed of the electrons, we can rearrange the formula to solve for \(v\): $$v = \sqrt{\frac{2K_e}{m}}$$ Plugging in the values, we get: $$v = \sqrt{\frac{2 \times 1.632 \times 10^{-18} J}{9.11 \times 10^{-31} kg}}$$ $$v \approx 7.66 \times 10^5 m/s$$

State the minimum speed of the electrons

The minimum speed of the electrons required to cause the emission of light due to the transition in the hydrogen atoms from \(n=2\) to \(n=1\) is approximately \(7.66 \times 10^5 \ m/s\).