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 frequency of the de-Broglie wave of an electron in Bohr's first orbit of a hydrogen atom is . . . . . . . . . . $\times 10^{13} \ Hz$ (nearest integer).
[Given: $R_H$ (Rydberg constant) $= 2.18 \times 10^{-18} \ J$,$h$ (Planck's constant) $= 6.6 \times 10^{-34} \ J \cdot s$]

The mass of an electron is $9.1 \times 10^{-31} \ kg$. If its $K.E.$ is $3.0 \times 10^{-25} \ J$,calculate its wavelength.

Calculate the $\lambda$ associated with an electron moving with the velocity of light.

What is the wavelength of an electron? What does it indicate?

If an electron,hydrogen nucleus,helium nucleus,and neon nucleus are all moving with the same velocity,then the wavelengths associated with these particles are in the order:

Vedclass Test Series

Exam Paper Generator

Online Exam Module