Question

In a photoelectric effect experiment, light from a blue LED (wavelength = 405 nm) is directed onto a piece of sodium. It is observed that when V = 0.52 volts, the current measured at the collector drops to zero. What is the work function of the sodium? (a) 0.85 eV (b) 2.5 eV (c) 1.6 eV (d) 3.6 eV (e) 0.088 eV 100% What is the highest velocity of an ejected electron, just above the surface of the sodium? (a) 680000 m/s (b) 300000 m/s (c) 430000 m/s 100% Now we decrease the power of the LED, keeping the wavelength fixed. Which of the following statements is true? (a) The maximum velocity of any ejected electrons will decrease. (b) The minimum possible time to eject an electron (relative to when the LED is turned on) will increase.

Answer 1

Answer:

1. (b) 2.5 eV

2. (c) 430000 m/s

3. (a) The maximum velocity of any ejected electrons will decrease.  

Explanation:

1)

From Einstein's Photoelectric Equation:

hc/λ = K.E + ∅

∅ = hc/λ - K.E

∅ = hc/λ - eV

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

V = Stopping Potential = 0.52 volts

h = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 405 nm = 4.05 x 10⁻⁷ m

∅ = Work Function = ?

Therefore,

∅ = (6.625 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4.05 x 10⁻⁷ m) - (1.6 x 10⁻¹⁹ C)(0.52 volts)

∅ = 4.9 x 10⁻¹⁹ J - 0.832 x 10⁻¹⁹ J

∅ = (4.075 x 10⁻¹⁹ J)(1 eV)/(1.6 x 10⁻¹⁹ J)

∅ = 2.5 eV

therefore, correct answer is"

(b) 2.5 eV

2)

K.E = (1/2)mv² = eV

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed = ?

therefore,

(1/2)(9.1 x 10⁻³¹ kg)v² = (1.6 x 10⁻¹⁹ C)(0.52 volts)

v = √(0.18 x 10¹² m²/s²)

v = 0.43 x 10⁶ m/s = 430000 m/s

Correct option is:

(c) 430000 m/s

3.

The decrease in power at constant wavelength means decrease in voltage, that results in the decrease of kinetic energy of electrons. So, correct option is:

(a) The maximum velocity of any ejected electrons will decrease.