I need help on this homework problem and I do not understand how to work on it. Please help me with this problem.
The photoelectric work function of a metal is the minimum energy needed to eject an electron by irradiating the metal with light. For calcium, this work function equals 4.34X10^(-19)J. What is the minimum frequency of light required to observe the photoelectric effect in calcium?
This question should have been posted on the PHYSICS board.
delta E = hf
Substitute 4.34 x 10^-19 for delta E, substitute Planck's constant for h and calculate frequency.
1.53e-15
To solve this problem, you need to use Einstein's photoelectric equation, which relates the energy (E) of a photon to its frequency (ν) and Planck's constant (h):
E = h * ν
In this equation, the energy (E) corresponds to the minimum energy needed to eject an electron from the metal (which is given as the photoelectric work function). The value of the photoelectric work function for calcium is given as 4.34 × 10^(-19) J.
We can rearrange the equation to solve for the frequency (ν):
ν = E / h
Now, substitute the given value of the photoelectric work function into the equation:
ν = 4.34 × 10^(-19) J / (6.63 × 10^(-34) J·s) [using the value of Planck's constant, h, as 6.63 × 10^(-34) J·s]
Now, perform the calculation:
ν ≈ 6.55 × 10^14 Hz
Therefore, the minimum frequency of light required to observe the photoelectric effect in calcium is approximately 6.55 × 10^14 Hz.