photons 2

Question

1- The energy of a photon of blue light that has a frequency of 7.50 x 10¹⁴ Hz is ___eV

2- Light with a wavelength of 425 nm falls on a photoelectric surface that has a work function of 2.0 eV. The maximum speed of any emitted photoelectrons is ___x10⁵ m/s

Answer 1

(1)

The frequency of light is f = 7.50 x 10¹⁴ Hz .

The Energy of photon can be found using the formula E_p = h*f .

Where h is plank's constant = 6.63 × 10^-34 J·sec .

            f is the frequency of light . 

 E_p = h*f ⇒ ( 6.63 × 10^-34 )*(7.50 x 10¹⁴)

            = 4.969 × 10^-19 .

The Energy of photon is 4.969 × 10^-19  eV .

Answer 2

(2)

Wave lenght of light  λ= 425 nm = 425 x 10^-9 m .

Work function w = 2 eV  = 3.2043 x 10^-19 J .

The kinetic energy of light E_k = E_p - w ⇒ [ (hc)/λ ] - w

Where h is plank's constant = 6.63 × 10^-34 J·sec .

          c = velocity of light = 3 × 10⁸ m/s .

E_k = [ (6.63 × 10^-34 × 3 × 10⁸ ) / 425 x 10^-9 ] - 3.2043 x 10^-19 J 

E_k = [ 4.6772 x 10^-19 ] - 3.2043 x 10^-19 J

E_k = 1.4729 x 10^-19 J .

Note : To determine the speed of the photoelectrons, you must use the kinetic energy expressed in joules .

The kinetic energy of light E_k = (1/2)m v² 

v = √ [ (2E_k)/m ]

Where m is mass of electron 9.11 x 10^-31  kg.

v = √ [ (2 x1.4729 x 10^-19) / 9.11 x 10^-31 ]

v = 5.686 x 10⁵  m/s .

Hence the speed of photo electrons is 5.686 x 10⁵ m/s .