what is potential difference for an electron to be accelerated to 0.25n m ?



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Posted on (Answer #1)

You need to use the equation that relates the kinetic energy and potential energy change, such that:

`(mv^2)/2 = eV`

Since the problem provides the wavelength, you may evaluate the speed v, using the following formula, such that:

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

`v = (6.63*10^(-34))/(0.24*10^(-9)*9.11*10^(-31))`

`v = 303*10^4 m/s`

Replacing `303*10^4` for v in equation `(mv^2)/2 = eV` yields:

`(9.11*10^(-31)*303^2*10^8)/2 = eV => eV = 419*10^(-16)` J

Hence, evaluating the potential difference yields `eV = ` `419*10^(-16)` J.

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