Question

Find the maximum Kinetic energy of emitted photoelectrons from metal surface when light of frequency 1 x 10^15 Hz is incident on it and the threshold wavelength for metal is 6000 A.U.(h = 6.625 x 10 ^ -34 J.s, c= 3 x 10 ^8 m/s)​

Answer 1

★$\color{darkblue}\underline{\underline{\sf Given-}}$

• Frequency of light (f) = ${\sf 10^{15}\:Hz}$
• Threshold Wavelength ${\sf (\lambda_o)=6000\:A°\: or \: 6×10^{-7}m}$
• Planck's constant (h) = ${\sf 6.625×10^{-34}Js}$
• Speed of light (c) = ${\sf 3×10^8\:m/s}$

★$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Maximum Kinetic Energy ${\sf (K_{max})}$

★$\color{darkblue}\underline{\underline{\sf Formula \: Used-}}$

$\color{violet}\underline{\boxed{\sf KE_{max}=E-\phi}}$

$\implies{\sf E =hf}$

$\implies{\sf \phi = \dfrac{hc}{\lambda_o}}$

★$\color{darkblue}\underline{\underline{\sf Solution-}}$

$\implies{\sf KE_{max}=hf-\dfrac{hc}{\lambda_o}}$

$\implies{\sf h(f-\dfrac{c}{\lambda_o} }$

$\implies{\sf 6.625×10^{-34}×\left(10^{15}-\dfrac{3×10^8}{6×10^{-7}}\right)}$

$\implies{\sf 6.625×10^{-34}×10^{15}-0.5×10^{15} }$

$\color{red}\implies{\sf KE_{max}=3.31×10^{-4}\:J}$

★$\color{darkblue}\underline{\underline{\sf Answer-}}$

Maximum Kinetic Energy is $\color{red}{\sf 3.31×10^{-4}J}$