The kinetic energy of the electron is E when the incident wavelength is λ . To increase the K . E. of the electron to 2 E, the incident wavelength must be

Question

The kinetic energy of the electron is E when the incident wavelength is λ . To increase the K . E. of the electron to 2 E, the incident wavelength must be (A) 2 λ (B) λ / 2 (C) (h c λ/E λ+h c) (D) (h c λ/E λ-h c)

Tardigrade Admin · Accepted Answer

(h c/λ)=W+E where E is the kinetic energy (h c/λ prime)=W+(2 E) ⇒ h c[(1/λ prime)-(1/λ)]=E (1/λ prime)=(E/h c)+(1/λ)=(E λ+h c/h c λ) ⇒ λ prime=(h c ⋅ λ/E λ+h c)

Q. The kinetic energy of the electron is $E$ when the incident wavelength is $λ .$ To increase the $K . E$ . of the electron to $2 E$ , the incident wavelength must be

$2 λ$

$λ /2$

$\frac{h c λ}{E λ + h c}$

$\frac{h c λ}{E λ - h c}$

$\frac{h c}{λ} = W + E$
where $E$ is the kinetic energy
$\frac{h c}{λ ^{'}} = W + (2 E) \Rightarrow h c [\frac{1}{λ ^{'}} - \frac{1}{λ}] = E$
$\frac{1}{λ ^{'}} = \frac{E}{h c} + \frac{1}{λ} = \frac{E λ + h c}{h c λ} \Rightarrow λ^{'} = \frac{h c \cdot λ}{E λ + h c}$

Q. The kinetic energy of the electron is E when the incident wavelength is λ. To increase the K.E. of the electron to 2E, the incident wavelength must be

2λ

λ/2

Eλ+hchcλ​

Eλ−hchcλ​