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calculate the wavelength in the visible spectrum (3500-7000)angstrom which coincide with the wavelength 6000 angstrom in 5th order?

Srishti Dubey , 5 Years ago
Grade 12th pass
anser 3 Answers
Pushpak

Last Activity: 4 Years ago

Since E=hv
And c= v× lambda
Therefore
E=hv = hc/ lambda = hc/6000 .     --->1
ZE= hv1 = hc/lambda                      ---->2
From   &.
Zhc/6000.  =  hc/lambda
Therefore
lambda1 = 3000 A°

Arun

Last Activity: 4 Years ago

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 

Vikas TU

Last Activity: 4 Years ago

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