A monochromatic light of wavelength λ ejects photoelectrons from a metal surface with work function ( varphi) 2.4 eV. These photoelectrons are made to collide with hydrogen atoms in ground state. The maximum value of λ for which hydrogen atom may be ionized is [Take hc=1240 eV-nm]

Question

A monochromatic light of wavelength λ ejects photoelectrons from a metal surface with work function ( varphi) 2.4 eV. These photoelectrons are made to collide with hydrogen atoms in ground state. The maximum value of λ for which hydrogen atom may be ionized is [Take hc=1240 eV-nm] (A) 80 nm (B) 77.5 nm (C) 75.5 nm (D) 85 nm

Tardigrade Admin · Accepted Answer

Given, work-function, φ=2.4 eV Energy required to ionise H-atom in ground state =13.6 eV According to photoelectric equation, E=K E+ varphi ⇒ (h c/λ)=K E+ varphi ⇒ (1240/λ)=13.6+2.4 (∵ h c=1240 eV - nm ) ⇒ (1240/λ)=16 ∴ λ=(1240/16) =77.5 nm

Q. A monochromatic light of wavelength λ ejects photoelectrons from a metal surface with work function (\varphi) 2.4eV. These photoelectrons are made to collide with hydrogen atoms in ground state. The maximum value of λ for which hydrogen atom may be ionized is [Take hc=1240{\,} eV-nm]

80 nm

77.5 nm

75.5 nm

85 nm

Given, work-function, ϕ=2.4eV Energy required to ionise H-atom in ground state =13.6eV According to photoelectric equation, E=KE+φ⇒λhc​=KE+φ ⇒λ1240​=13.6+2.4 (∵hc=1240eV−nm) ⇒λ1240​=16 ∴λ=161240​ =77.5nm

Q. A monochromatic light of wavelength $λ$ ejects photoelectrons from a metal surface with work function (\varphi) $2.4 e V$ . These photoelectrons are made to collide with hydrogen atoms in ground state. The maximum value of $λ$ for which hydrogen atom may be ionized is [Take hc=1240{\,} eV-nm]