The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength :

Question

The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength : (A) (h c/E λ-h c) (B) (h c λ/E λ+h c) (C) ( h λ/ E λ+ hc ) (D) (h c λ/E λ-h c)

Tardigrade Admin · Accepted Answer

E =( hc /λ)-φ-( i ) 2 E =( hc /λ prime)-φ-( text ii) (ii) - (i) E = hc ((1/λ prime)-(1/λ)) ⇒ λ prime=( hc λ/ E λ+ hc )

Q. The kinetic energy of emitted electron is $E$ when the light incident on the metal has wavelength $λ$ . To double the kinetic energy, the incident light must have wavelength :

$\frac{h c}{E λ - h c}$

$\frac{h c λ}{E λ + h c}$

$\frac{hλ}{E λ + h c}$

$\frac{h c λ}{E λ - h c}$

$E = \frac{h c}{λ} - ϕ - (i)$
$2 E = \frac{h c}{λ ^{'}} - ϕ - (ii)$
(ii) - (i)
$E = h c (\frac{1}{λ ^{'}} - \frac{1}{λ})$
$\Rightarrow λ^{'} = \frac{h c λ}{E λ + h c}$

Q. The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength :

Eλ−hchc​

Eλ+hchcλ​

Eλ+hchλ​

Eλ−hchcλ​