Question

The energy needed to ionize an atom of si when it is in the most stable is 786.4 kJ mol^-1 however if an atom of Si is in certain low lying excited state only 310.8 is needed to ionize.
what is the wavelength of he radiation emitted when an atom of si undergoes a transition from this excited state to the ground state?

Answer 1

Answer:

The wavelength of he radiation emitted is $\lambda = 252 \ nm$

Explanation:

We know that energy needed is given by

$E = \frac{h c}{\lambda}$

$\lambda = \frac{hc}{E}$  ----- (1)

h = 6.62 × $10^{-34}$ J s

c = 3  × $10^{8} \ \frac{m}{s}$

E = 786.4 - 310.8 = 475.6 $\frac{KJ}{mole}$

Total energy

$E = \frac{475.6(10^{3} )}{(6.023)10^{23} }$

E = 78.9 × $10^{-20}$ J

From equation (1)

$\lambda = \frac{6.626(10^{-34} )3 (10^{8} )}{78.9 (10^{-20} )}$

$\lambda = 2.52$ × $10^{-7}$

$\lambda = 252 \ nm$

Therefore the wavelength of he radiation emitted is $\lambda = 252 \ nm$