Question

The kinetic energy of the electron orbiting in the first excited state of a hydrogen atom is 3.4 eV. Determine the de Broglie wavelength associated with it.

(a) λ=h/√2mE​​
(b) λ=​h/√mE​
(c) λ=​h/√2m​
(d) λ=​h/√2mE​

Answer 1

To calculate the de Broglie wavelength of an electron with a kinetic energy of 3.4 eV, use the formula λ = h / √(2mE) and convert the energy from eV to J before substituting the values.

Step-by-step explanation:

The kinetic energy of the electron orbiting in the first excited state of a hydrogen atom is 3.4 eV. To determine the de Broglie wavelength associated with it, we can use the formula λ = h / √(2mE), where λ is the wavelength, h is Planck's constant (6.626×10⁻³´ Js), m is the mass of the electron (9.109×10⁻¹¹ kg), and E is the kinetic energy of the electron in joules.

To convert the kinetic energy from electron volts (eV) to joules (J), we use the conversion factor 1 eV = 1.602×10⁻¹⁹ J. Therefore:

E (in J) = 3.4 eV × 1.602×10⁻¹⁹ J/eV = 5.447×10⁻¹¹ J

Using the formula for the de Broglie wavelength, we have:

λ = h / √(2mE) = (6.626×10⁻³´ Js) / √(2 × 9.109×10⁻¹¹ kg × 5.447×10⁻¹¹ J)

After calculating, we can find the de Broglie wavelength of the electron. In this case, the correct formula to use is option (d): λ = h / √(2mE).