Answer:
E = 45.08×10⁻¹⁹j
Explanation:
Given data:
Energy of photon in joule = ?
Frequency of photon = 6.8×10¹⁵ Hz
Solution:
Formula:
E = h.f
f = frequency
h = planck's constant (6.63×10⁻³⁴Js)
E = energy
Value of planck's constant is 6.63×10⁻³⁴Js.
E = 6.63×10⁻³⁴Js ×6.8×10¹⁵ Hz
Hz = s⁻¹
E = 6.63×10⁻³⁴Js ×6.8×10¹⁵ s⁻¹
E = 45.08×10⁻¹⁹j
Energy is the quantitative unit, which is measured in joules. The energy of a photon of UV light having the frequency 6.8×10¹⁵ Hz is 45.08×10⁻¹⁹ Joules.
Given:
Frequency of photon = 6.8×10¹⁵ Hz
Energy of a photon in joule =?
Usingt the formula of energy:
E = h x v
where,
v = frequency
h = planck's constant (6.63×10⁻³⁴Js)
E = energy
Substituting the values in the equation, we get:
E = 6.63×10⁻³⁴Js ×6.8×10¹⁵ Hz
E = 6.63×10⁻³⁴Js ×6.8×10¹⁵ s⁻¹
E = 45.08×10⁻¹⁹ Joules
Therefore, the energy of a photon in the UV region is 45.08×10⁻¹⁹ Joules
To know more about energy, refer to the following link:
https://brainly.com/question/5121707
