## Physics Planck Equation Question

(c) Tenaga minimum yang diperlukan untuk fotoelektron terlepas dari permukaan logam litium ialah 2.32 \mathrm{eV}.

The minimum energy required for a photoelectron to escape from the lithium metal surface is 2.32 \mathrm{eV}.(i) Logam litium tidak menunjukkan kesan fotoelektrik untuk cahaya biru yang mempunyai panjang gelombang 460 \mathrm{~nm}. Terangkan. Lithium metal does not show a photoelectric effect for blue light with a wavelength of 460 \mathrm{~nm}. Explain.

(ii) Berapakah panjang gelombang ambang bagi litium? What is the threshold wavelength for lithium?

### Physics Planck Equation Solution

**Part c(ii)**

First, we need to identify this question is from which Physics topic. Looking at keywords like **“photoelectric”**, **“energy”** and **“wavelength”**, we know it is certainly related to Planck’s equation.

Now let’s refer to Planck’s equation definition:

The Planck relation states that the energy of a photon, E, known as photon energy, is proportional to its frequency, ν: The constant of proportionality, h, is known as the Planck constant.

Its formula is:

E = hv = \frac{hc}{\lambda}

where h = Planck Constant, c = speed of light, \lambda = wavelength, E=energy

From the question, we can extract following information:

Minimum Energy for a photoelectron to escape, E = 2.32 eV

Note 1eV = 1.6 \times 10^{-19}V,

Hence, E = 2.32 eV = 2.32 \times 1.6 \times 10^{-19}

From formula sheet,

Planck Constant, h = 6.626 \times 10^{-34}

Speed of light, c = 3.0 \times 10^8

Plugging all the above values into Planck’s equation,

(2.32 \times 1.6 \times 10^{-19}) = \frac{(6.626 \times 10^{-34}) \times ( 3.0 \times 10^8)}{\lambda}

\lambda = 5.355 \times 10^{-7} m

\lambda = 535.5 \times 10^{-9} m

\lambda = 535.5 nm