Question

if the energy difference between the ground state and excited state of an atom is 4.4×10^-19J . the wavelength of photon required to produce this transition​

Answer 1

Answer:

wavelength =4.5×10^7

Explanation:

energy difference = hc/wavelength

for detailed explanation view attachment.

Answer 2

Answer: $4.5\times 10^{-7}m$

Explanation:

$\Delta E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation  = ?

h= plank's connstant =$6.6\times 10^{-34}sec^{-1}$

c = velocity of light =$3\times 10^8ms^{-1}$

$\Delta E$= energy  difference = $4.4\times 10^{-19}J$

Putting in the values we get,

$4.4\times 10^{-19}=\frac{6.6\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=4.5\times 10^{-7}m$

The wavelength of photon required to produce this transition​ is $4.5\times 10^{-7}m$