Answer:
The wavelength of he radiation emitted is [tex]\lambda = 252 \ nm[/tex]
Explanation:
We know that energy needed is given by
[tex]E = \frac{h c}{\lambda}[/tex]
[tex]\lambda = \frac{hc}{E}[/tex] ----- (1)
h = 6.62 × [tex]10^{-34}[/tex] J s
c = 3 × [tex]10^{8} \ \frac{m}{s}[/tex]
E = 786.4 - 310.8 = 475.6 [tex]\frac{KJ}{mole}[/tex]
Total energy
[tex]E = \frac{475.6(10^{3} )}{(6.023)10^{23} }[/tex]
E = 78.9 × [tex]10^{-20}[/tex] J
From equation (1)
[tex]\lambda = \frac{6.626(10^{-34} )3 (10^{8} )}{78.9 (10^{-20} )}[/tex]
[tex]\lambda = 2.52[/tex] × [tex]10^{-7}[/tex]
[tex]\lambda = 252 \ nm[/tex]
Therefore the wavelength of he radiation emitted is [tex]\lambda = 252 \ nm[/tex]
