# IBDP Physics E.2 Quantum physics IB Style Question Bank : HL Paper 1

### Question

What is the pattern observed when white light passes through a diffraction grating? Ans:C

When white light passes through a diffraction grating, a pattern known as a “spectral or diffraction pattern” is observed. This pattern is a result of the dispersion of light into its component colors due to the phenomenon of diffraction and interference. ### Question

The energy levels E of an atom are shown. Which emission spectrum represents the transitions? Ans:D

### Question

Some energy levels for a hydrogen atom are shown. What is the $$\frac{\text { wavelength emitted in transition } \mathrm{X}}{\text { wavelength emitted in transition } \mathrm{Y}}$$ ?

A. $$\frac{1}{2}$$

B. $$\frac{27}{32}$$

C. $$\frac{32}{27}$$

D. 2

Ans:C

The wavelength of light emitted in a hydrogen atom transition can be calculated using the Rydberg formula:

$$\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)$$

For transition X (from $$n=2$$ to $$n=1$$), we have:

$$\frac{1}{\lambda_X} = R_H \left(\frac{1}{1^2} – \frac{1}{2^2}\right) = R_H \left(1 – \frac{1}{4}\right) = R_H \cdot \frac{3}{4}$$

$\lambda_X=(\frac{1}{R_H})\times \frac{4}{3}$

For transition Y (from $$n=3$$ to $$n=1$$), we have:

$$\frac{1}{\lambda_Y} = R_H \left(\frac{1}{1^2} – \frac{1}{3^2}\right) = R_H \left(1 – \frac{1}{9}\right) = R_H \cdot \frac{8}{9}$$

$\lambda_Y=(\frac{1}{R_H})\times \frac{9}{8}$

So, the $$\frac{\text{wavelength emitted in transition X}}{\text{wavelength emitted in transition Y}}=\frac{32}{27}$$

### Question

Light of frequency $$f$$ is incident on a metallic surface of work function $$W$$. Photoelectrons with a maximum kinetic energy $$E_{\max }$$ are emitted. The frequency of the incident light is changed to $$2 f$$.
What is true about the maximum kinetic energy and the work function? Ans:C

The maximum kinetic energy of emitted photoelectrons is determined by the energy of the incident photons and the work function of the material. The work function is the minimum energy required to release an electron from the metal surface. The maximum kinetic energy ($$E_{\max}$$) of emitted photoelectrons is given by the equation:

$E_{\max} = hf – W$

If the frequency of the incident light is changed to $$2f$$, the new maximum kinetic energy ($$E’_{\max}$$) of the emitted photoelectrons will be:

$E’_{\max} = 2hf – W$

Comparing these two equations, we can see that $$E’_{\max}$$ is greater than $$2E_{\max}$$ because $$2hf$$ is greater than $$hf$$, and $$E’_{\max}$$ is reduced by the same work function $$W$$.

So, the correct option is:

greater than $$2E_{\max}$$ & unchanged

### Question

What is not correct about a photovoltaic cell?

A. It has an output power that is related to the surface area of the cell.

B. It generates an alternating current.

C. It absorbs energy over a range of photon frequencies.

D. It can be used to store energy in a secondary cell.