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

Pushpak
15 Points
3 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
25757 Points
3 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

$c= \nu \lambda$

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
14149 Points
3 years ago
Dear student
The above explanation is incorrect , Please do not follow..................................................
Good Luck