An experiment was carried out. When radiating with a frequency of 5.5 * 10 ^ 4 Hz, the cut-off voltage was 0.4 V

An experiment was carried out. When radiating with a frequency of 5.5 * 10 ^ 4 Hz, the cut-off voltage was 0.4 V, at a frequency of 6.1 * 10 ^ 14 Hz, the cut-off voltage was 0.6 V. What is Planck’s constant?

Let’s write down the Einstein equation for both cases (and compose a system of them)
hv1 = Aout + (mv1 ^ 2) / 2
hv2 = Aout + (mv2 ^ 2) / 2

Let’s subtract them from each other:
h (v1 – v2) = (mv1 ^ 2) / 2 – (mv2 ^ 2) / 2

We write down the kinetic energy, we express Planck’s constant h:
Wк = | Uз * q |

h = (| Uz1 * q | / * | Uz2 * q |) / (v1 – v2)
h = (q * | Uz1 – Uz2 |) / (v1- v2)

Substitute the numerical values:
h = [1.6 * 10 ^ -19 * (0.4 – 0.6)] / [(5.5 – 6.1) * 10 ^ 14]
h = 0.32 / 0.6 * 10 ^ 33
h = 5.33 * 10 ^ -34

Answer: h = 5.33 * 10 ^ -34