I need help with this problem so please help me.
1. A certain photoelectric surface has a threshold frequency of 6.2*10^14 Hz.
A. How much kinetic energy is required to produce an electron emission?
B. if the surface is illuminated with light that has a frequency 7.1*10^14 Hz, what is the kinetic energy of the emitted electron in joules?
C. What is the kinetic energy of the emitted electron in electron-volts?
Sure! I can help you with these questions.
A. To find the kinetic energy required to produce an electron emission, we can use the equation:
E = hf - Φ
where E is the kinetic energy, h is the Planck's constant (6.62607004 × 10^-34 J·s), f is the frequency of the incident light, and Φ is the threshold frequency.
In this case, the threshold frequency is given as 6.2*10^14 Hz. So, we can substitute the values into the equation:
E = hf - Φ
E = (6.62607004 × 10^-34 J·s) * (6.2*10^14 Hz) - (6.2*10^14 Hz)
E = 4.10621 × 10^-19 J
Therefore, the kinetic energy required to produce an electron emission is approximately 4.10621 × 10^-19 Joules.
B. To find the kinetic energy of the emitted electron when illuminated with light of frequency 7.1*10^14 Hz, we can use the same equation as before:
E = hf - Φ
Substituting the values:
E = (6.62607004 × 10^-34 J·s) * (7.1*10^14 Hz) - (6.2*10^14 Hz)
E = 7.834659844 × 10^-19 J
So, the kinetic energy of the emitted electron when illuminated with light of frequency 7.1*10^14 Hz is approximately 7.834659844 × 10^-19 Joules.
C. To convert the kinetic energy from Joules to electron-volts (eV), we can use the conversion factor: 1 eV = 1.602176634 × 10^-19 J.
So, to find the kinetic energy in electron-volts, we can divide the kinetic energy in Joules by this conversion factor:
Kinetic energy in eV = (Kinetic energy in J) / (1.602176634 × 10^-19 J)
Kinetic energy in eV = (7.834659844 × 10^-19 J) / (1.602176634 × 10^-19 J)
Kinetic energy in eV ≈ 4.88538957 eV
Therefore, the kinetic energy of the emitted electron in electron-volts is approximately 4.88538957 eV.