An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is :

Question

An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is : (A) ( (E/2m) )(1/2) (B) c(2mE )(1/2) (C) (1/c) ( (2m/E) )(1/2)c (D) (1/c) ( (E/2m) )(1/2)c

Tardigrade Admin · Accepted Answer

For electron λe = (h/√2mE) for Photon E = pc ⇒ λPh = (hc/E) ⇒ (λe/λPh) = (h/√2mE) × (E/hc) = ( (E/2m) )1/2 (1/c)

Q. An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is :

$(\frac{E}{2 m})^{\frac{1}{2}}$

$c (2 m E)^{\frac{1}{2}}$

$\frac{1}{c} (\frac{2 m}{E})^{\frac{1}{2}}$ c

$\frac{1}{c} (\frac{E}{2 m})^{\frac{1}{2}}$ c

For electron $λ_{e} = \frac{h}{2 m E}$
for Photon $E = p c \Rightarrow λ_{P h} = \frac{h c}{E}$
$\Rightarrow \frac{λ _{e}}{λ _{P h}} = \frac{h}{2 m E} \times \frac{E}{h c} = (\frac{E}{2 m})^{1/2} \frac{1}{c}$

Q. An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is :

(2mE​)21​

c(2mE)21​

c1​(E2m​)21​c

c1​(2mE​)21​c

For electron λe​=2mE​h​ for Photon E=pc⇒λPh​=Ehc​ ⇒λPh​λe​​=2mE​h​×hcE​=(2mE​)1/2c1​

$(\frac{E}{2 m})^{\frac{1}{2}}$

$c (2 m E)^{\frac{1}{2}}$

$\frac{1}{c} (\frac{2 m}{E})^{\frac{1}{2}}$ c

$\frac{1}{c} (\frac{E}{2 m})^{\frac{1}{2}}$ c

For electron $λ_{e} = \frac{h}{2 m E}$
for Photon $E = p c \Rightarrow λ_{P h} = \frac{h c}{E}$
$\Rightarrow \frac{λ _{e}}{λ _{P h}} = \frac{h}{2 m E} \times \frac{E}{h c} = (\frac{E}{2 m})^{1/2} \frac{1}{c}$