Answer:
The value is [tex]E_i = 1.5596 *10^{-18} \ J[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m[/tex]
The velocity is [tex]v = 2.371*10^6 \ m/s[/tex]
The mass of electron is [tex]m_e = 9.109*10^{-31} \ kg[/tex]
Generally the energy of the incident light is mathematically represented as
[tex]E = \frac{h * c}{\lambda}[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
h is the Planck constant with value [tex]h = 6.62607015 * 10^{-34 } J\cdot s[/tex]
So
[tex]E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}[/tex]
=> [tex]E = 4.12 *10^{-18} \ J [/tex]
Generally the kinetic energy is mathematically represented as
[tex]E_k = \frac{1}{2} * m_e * v^2[/tex]
=> [tex]E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2[/tex]
=> [tex]E_k = 2.56 *0^{-18} \ J [/tex]
Generally the ionization energy is mathematically represented as
[tex]E_i = 4.12 *10^{-18} - 2.56 *0^{-18}[/tex]
=> [tex]E_i = 1.5596 *10^{-18} \ J[/tex]