The wavelength of a photon is 5200 mathring{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.For an electron,the velocity $v$ is given by $v

Vedclass · Accepted Answer

$1400$

Vedclass · Answer

(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.For an electron,the velocity $v$ is given by $v = \frac{h}{m \lambda}$.Given: $h = 6.626 	imes 10^{-34} \ J \cdot s$,$m = 9.11 	imes 10^{-31} \ kg$,and $\lambda = 5200 \ \mathring{A} = 5.2 	imes 10^{-7} \ m$.Substituting the values:$v = \frac{6.626 	imes 10^{-34}}{(9.11 	imes 10^{-31}) 	imes (5.2 	imes 10^{-7})}$$v = \frac{6.626 	imes 10^{-34}}{47.372 	imes 10^{-38}}$$v = 0.1398 	imes 10^4 \ m/s \approx 1400 \ m/s$.

The wavelength of a photon is $5200 \ \mathring{A}$. At what velocity will an electron have the same wavelength as this photon (in $m/s$)?

Similar Questions

The wavelengths of electron waves in two orbits are in the ratio $3: 5$. The ratio of the kinetic energy of the electrons will be:

$A$ ball weighing $25 \, g$ moves with a velocity of $6.6 \times 10^4 \, cm/sec$. The de Broglie wavelength associated with it is .....

The de Broglie equation applies to:

If a moving proton,deuteron,and alpha particle have the same De Broglie wavelength,then what is the correct order of their kinetic energy?

The wavelength of an electron in the first orbit of a hydrogen atom is $3.3 \times 10^{-10} \ m$. The kinetic energy of the electron (in $J$) is:
$(h = 6.6 \times 10^{-34} \ Js, m_{e} = 9.0 \times 10^{-31} \ kg)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module