Question

What is the de Broglie wavelength of an electron traveling at 1.62×105 m/s ?

Answer 1

Answer : The wavelength of an electron is,
`4.39* 10^(-9)m`

Explanation :

According to de-Broglie, the expression for wavelength is,

`\lambda=(h)/(p)`

and,

`p=mv`

where,

p = momentum, m = mass, v = velocity

So, the formula will be:

`\lambda=(h)/(mv)`

where,

h = Planck's constant =
`6.626* 10^(-34)Js`

`\lambda` = wavelength = ?

m = mass of electron =
`9.31* 10^(-31)kg`

v = velocity =
`1.62* 10^5m/s`

Now we have to calculate the wavelength.

Now put all the given values in the above formula, we get:

`\lambda=(h)/(mv)`

`\lambda=(6.626* 10^(-34)Js)/((9.31* 10^(-31)kg)* (1.62* 10^5m/s))`

`\lambda=4.39* 10^(-9)m`

Therefore, the wavelength of an electron is,
`4.39* 10^(-9)m`

Answer 2

0.449 10^(-8) meters

`0.449 * 10^(-8)` meters

Step-by-step explanation:

Use the deBroglie equation for the wavelength (lambda):

`lambda = (h)/(m*v)`

where h stands for the Plank constant:
`6.6261 * 10^(-34) J`

m stands for the mass of the electron:
`9.109 * 10^(-31) kg`

and v is the given velocity:
`v = 1.62 *  10^5 (m)/(s)`

Evaluating lambda for these values:

`lambda = (h)/(m*v)= (6.6261 * 10^(-34) )/(9.109 * 10^(-31) * 1.62 * 10^(5)) = 0.449* 10^(-8) m`