Question

What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect?

Answer 1

Answer:

1.05 × 10 ∧15 Hz.

Explanation:

The minimum amount of energy of photon E = 6.94×10∧-19  J

The value of Plank's constant , h = 6.626 × 10∧-34 J s

So the minimum frequency of light necessary to emit electrons from titanium via photoelectric effect , E = h . ν

where ν is the frequency

                                       ν = E / h

                                       ν = 6.94×10∧-19  J / 6.626 × 10∧-34 J s

                                         = 1.05 × 10 ∧15 / s

                                         = 1.05 × 10 ∧15 Hz.

Answer 2

E = h x f 

. . . then : 

f = E / h 
f = 4,41•10^-19 / 6,62•10^-34 
f = 6,66•10^14 Hz (s^-1) 


b/ What is the wavelength of this light ? 
- - - - - - - - - - - - - - - - - - - - - - - - - - - - 

λ = c / f 
λ = 3•10^8 / 6,66•10^14 
λ = 4,50•10^-7 m