A photon of wavelength 6000 Å has an energy E. The wavelength of the photon of light which corresponds to an energy equal to 2E is

  • Option 1)

    300 Å

  • Option 2)

    3000 Å

  • Option 3)

    30 Å

  • Option 4)

    3 Å

 

Answers (1)
V Vakul

As learnt

The energy (E) of a quantum of radiation -

E=hv

Where h is plank’s constant and \nu is frequency

-

 

 AND
 

 

Speed of electromagnetic radiation -

c= \nu \lambda

where frequency is (\nu,pronounced as nu), wavelength is (λ) and velocity of light is(c)

 

-

 

 We know, E=h\nu

E= energy of a quantum of radiation

h=plank's constant

\nu= frequency of radiation

and c=\nu\lambda

Now, E=h\nu=\frac{hc}{\lambda }=\frac{hc}{6000} -(1)

ZE=hv_{1}=h\frac{c}{\lambda _{1}} - (2)

Put in 2

Z\times \frac{hc}{6000A}=\frac{hc}{\lambda_{1} }

\lambda _{1}=3000A 

 


Option 1)

300 Å

Option is incorrect

Option 2)

3000 Å

Option is correct

Option 3)

30 Å

Option is incorrect

Option 4)

3 Å

Option is incorrect

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