Question

Find the kinetic energy of an electron (in eV) emitted by a metal when light of wavelength 3105 angstrom falls on it
threshold wavelength is 6210 angstrom)​

Answer 1

Given :-

◉ Wavelength of light that falls on the metal, λ₁ = 3105 Å

◉ Threshold wavelength, λ = 6210 Å

To Find :-

⇒ Kinetic energy of the electron emitted (in eV)

Solution :-

We know, Energy of the photon of wavelength 3105 Å

⇒ E = hv

Where,

• v = frequency

⇒ E = h c/λ

⇒ E = 6.6×10⁻³⁴ × 3 × 10⁸ / 3105 × 10⁻¹⁰

⇒ E = 19.8 × 10⁻²⁶ / 3.1 × 10⁻⁷

⇒ E = 6.38 × 10⁻¹⁹ Joule

Now, We need to calculate it in eV , but we know

⇒ 1 Joule = 6.242 × 10¹⁸ eV

⇒ E = 6.38 × 10⁻¹⁹ × 6.242 × 10¹⁸ eV

⇒ E = 1.02 × 10⁻¹ eV

⇒ E = 0.1 eV

Hence, The kinetic energy of the electron is 0.1 eV.

Some Information :-

◉ The threshold wavelength or frequency of a metal is the minimum wavelength or frequency which a photon is required to have to eject electrons after falling on a metal.

• Any wavelength or frequency of photon below the threshold wavelength or frequency won't eject electrons.