The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength of the incident radiation is changed from 400 nm to 310 nm . The work function of the metal is x eV . Find 10x (given hc=1240 eV nm )

Question

The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength of the incident radiation is changed from 400 nm to 310 nm . The work function of the metal is x eV . Find 10x (given hc=1240 eV nm ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Energy of photon of 400 nm wavelength =(1240/400)=3.1 eV Energy of photon of 310 nm wavelength =(1240/310)=4 eV hv=φ+kmax⇒ 3.1=φ+kmax 4=φ+2kmax ⇒ φ=2.2 eV

Q. The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength of the incident radiation is changed from $400 nm$ to $310 nm$ . The work function of the metal is $x e V$ . Find $10 x$ (given $h c = 1240 e V nm$ )

Energy of photon of $400 nm$ wavelength $= \frac{1240}{400} = 3.1 e V$
Energy of photon of $310 nm$ wavelength $= \frac{1240}{310} = 4 e V$
$h v = ϕ + k_{ma x} \Rightarrow 3.1 = ϕ + k_{ma x}$
$4 = ϕ + 2 k_{ma x}$
$\Rightarrow ϕ = 2.2 e V$

Q. The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength of the incident radiation is changed from 400nm to 310nm . The work function of the metal is xeV . Find 10x (given hc=1240eVnm )

Energy of photon of 400nm wavelength =4001240​=3.1eV Energy of photon of 310nm wavelength =3101240​=4eV hv=ϕ+kmax​⇒3.1=ϕ+kmax​ 4=ϕ+2kmax​ ⇒ϕ=2.2eV

Q. The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength of the incident radiation is changed from $400 nm$ to $310 nm$ . The work function of the metal is $x e V$ . Find $10 x$ (given $h c = 1240 e V nm$ )

Energy of photon of $400 nm$ wavelength $= \frac{1240}{400} = 3.1 e V$
Energy of photon of $310 nm$ wavelength $= \frac{1240}{310} = 4 e V$
$h v = ϕ + k_{ma x} \Rightarrow 3.1 = ϕ + k_{ma x}$
$4 = ϕ + 2 k_{ma x}$
$\Rightarrow ϕ = 2.2 e V$