In a photoelectric experiment using a sodium surface, you find a stopping potential of 1.61 v for a wavelength of 320 nm and a stopping potential of 0.740 v for a wavelength of 412 nm. from these data find (a) a value for the planck constant, (b) the work function φ for sodium, and (c) the cutoff wavelength λ0 for sodium.

a) 4.15×10∧-15 eV/s

b) 1.31 eV

c) 950.4 nm

Explanation:

let Ф be the work function.

a) f = c/λ = (3×10^8)/(320×10^-9) = 9.375×10^14 /s

  f = c/λ = (3×10^8)/(412×10^-9) = 7.28×10^14 /s

E = hf +Ф

Energy which relates to the poteial is calculated in eV (electonvolts).

so E = eV

Now using the straight line analogy in the equation above h is the gradient of the line.

Note that energy which relates to the poteial is calculated in eV (electonvolts).

so E = eV

and h = eV/f

h=[(0.74 - 1.61)eV]/[((7.28 - 9.375)×10^14 )/s]

 = 4.15×10^-15 eV/s ≈ 6.65×10^-34 J/s

b)  E = hf +Ф

  eV = hc/λ - Ф

Ф = hc/λ -  eV

   = (4.15×10^-15 eV/s)(3×10^8)/[(320×10^-9) -  (1.60×10^-19)(1.61)eV]

   = 1.31 eV ≈ 2.1×10^-19 J/s

c) Ф = hc/λ0

  λ0 = hc/Ф

       = [(6.65×10^-34 J/s)(3×10^8 )]/ (2.1×10^-19 J/s)

       = 9.50×10^-7 m

the amount of current in the coil

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000