Question

2. Determine the energy of visible light (in eV) with a frequency of 7.5x1014 Hz
need solution and the answer sould not be round off at the middle of the computation.

Answer 1

Answer:

ENERGY(E) = PLANKS CONSTANT(h) x FREQUENCY (μ)

PLANKS CONSTANT(h) = 6.626 x 10⁻³⁴ Js

FREQUENCY(μ) = 7.5 x 10¹⁴ Hz

= E = hμ

= E = 6.626 x 10⁻³⁴ x 7.5 x 10¹⁴ Js x 1/s

= E = 49.6950 x 10⁻²⁰ J

FOR THE CONVERSION OF JOULES TO ENERGY

1000 J = 6.242 x 10²¹ eV

SO 49.695 x 10⁻²⁰ J = ? eV

BY THE MEANS OF CROSS MULTIPLICATION WE GET

 = (6.242 x 10²¹ x 49.695 x 10⁻²⁰ )/10³

 =  WE CAN APPROXIMATE THE ENERGY VALUE TO 50 FOR EASY CALCULATION

 = 6.242 x 50 x 10⁻²

 = 31.210 x 10⁻¹ eV

SO THE ENERGY OF THE VISIBLE LIGHT IS 31.210 x 10⁻¹ eV

Step-by-step explanation:

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Answer 2

Given:

Frequency of visible light = 7.5 × 10¹⁴ Hz

To find:

Energy of visible light

Solution:

When electromagnetic radiation hits a material, electrons are emitted. This phenomenon is called photoelectric effect. The electrons emitted in this way is called photoelectrons or photons. These photons contain a fixed amount of energy that depends on the frequency of the electromagnetic radiation.

Let $E$ be the amount of energy that the photons contain and $f$ be the frequency of the radiation, then

$E=hf$

where $h$ is Planck's constant and $h=6.62$ × $10^{-34}m^{2}kg/s$

Here, $f=7.5$ × $10^{14}Hz$     (given)

Substituting in the formula,

$E = 6.62*10^{-34}*7.5*10^{14}$

$E=6.62*7.5*10^{14-34}$

$E=49.65$ × $10^{-20}J$

To convert into electron volt (eV),

$1J=6.24$  × $10^{18}eV$

Multiply the energy obtained with 1 Joule amount to result the energy in eV.

$E=49.65*10^{-20}*6.24*10^{18}$

$E=49.65*6.24*10^{18-20}$

$E=309.816$ ×$10^{-2}=3.09eV$

3.09 eV is the amount of energy of visible light having frequency 7.5×10¹⁴Hz