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V125BC [204]
2 years ago
11

What is the frequency of a photon with an energy of 4. 56 x 10^-19 j

Physics
2 answers:
Sauron [17]2 years ago
5 0

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

#SPJ4

love history [14]2 years ago
3 0

Answer:

ν = E/h = 4.56x10-19 J / 6.626x10-34 Jsec-1

ν = 6.88x1014 s-1

Explanation:

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Answer:

a

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b

   The resultant displacement in the north east direction is

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From the question we are told that  

  The displacement in the morning is  S_{morning} =  (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)

   The displacement in the afternoon is  S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)

Generally the direction west is negative , the direction east is positive

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resultant displacement  is mathematically evaluated as  

    (2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east

     (4000\ m\ north)  + (2500 \ m ,\ north) = 6500  \ m ,\ north

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From the above calculation we see that at the end of the hiking the hiker (you) is  200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

      a  =  \sqrt{900^2 + 6500^2}

=>     a  = 6562.0 \  m

Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

      b = \sqrt{6500^2 +(-200)^2  }

=>   b =6503.1 \  m

   

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