Answer:
No photoelectric effect is observed for Mercury.
Explanation:
From E= hf
h= Plank's constant
f= frequency of incident light
Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23
f= 11×10^14 Hz
The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface
Answer:
Faraday's constant will be smaller than it is supposed to be.
Explanation:
If the copper anode was not completely dry when its mass was measured, mass of the copper must be heavier than it should have been. Hence, the calculated Faraday’s constant would be smaller than it is supposed to be since when calculating Faraday’s Constant, the charge transferred is divided by the moles of electrons.
Answer:
v = 7.3 × 10⁶ m/s
Explanation:
Given data:
Velocity of electron = ?
Wavelength = 100 pm
Solution:
Formula:
λ = h/mv
λ = wavelength
h = planck's constant
m = mass
v = velocity
Now we will put the values in formula.
100 ×10⁻¹² m = 6.63 × 10⁻³⁴ j.s / 9.109 × 10⁻³¹ kg × v
v = 6.63 × 10⁻³⁴ kg.m²/s / 9.109 × 10⁻³¹ kg ×100 ×10⁻¹² m
v = 6.63 × 10⁻³⁴ m/s /910.9 × 10⁻⁴³
v = 0.0073 × 10⁹ m/s
v = 7.3 × 10⁶ m/s
