An electron is continuonsly accelerated...

An electron is continuonsly accelerated in vacume tube by appliying potential difference if its de Brogle wavelength is decresed by `1%` the change in the kinetic energy of the electron is nearly

Decreased by `1%`

Increased by `2%`

Increased by `1%`

Decreased by `2%`

Text Solution

Verified by Experts

The correct Answer is:

`(hc)/(lambda)=E_(1)-E_(2)=KE_(2)-KE_(1)`
`:. Lambda=h/(mV) (mV)^(2)=(h/lambda)^(2), 1/2 mV^(2)=1/(2m)h^(2)/lambda^(2)`
`:. (hc)/lambda=h^(2)/(2m lambda_(2)^(2))-h^(2)/(2mlambda_(1)^(2)) :. Lambda=(2mc)/(h){(lambda_(1)^(2)lambda_(2)^(2))/(lambda_(1)^(2)-lambda_(2)^(2))}`

Topper's Solved these Questions

NUCLEAR CHEMISTRY
RESONANCE|Exercise Exercise-23|1 Videos
NUCLEAR CHEMISTRY
RESONANCE|Exercise Exercise-24|1 Videos
NUCLEAR CHEMISTRY
RESONANCE|Exercise Exercise-21|1 Videos
NITROGEN CONTAINING COMPOUNDS
RESONANCE|Exercise ORGANIC CHEMISTRY(Nitrogen containing Compounds)|30 Videos
P BLOCK ELEMENTS
RESONANCE|Exercise PART -II|24 Videos

An electron is continuously accelerated in a vacuum tube by applying potential differece. If the de-Broglie's wavelength is decreased by 10% , the change in the kinetic energy of the electron is nearly

decreased by `11%`

increased by `23.4%`

increased by `10%`

increased by `11.1%`

An electron and a proton are accelerated through the same potential difference. The ratio of their de-Broglie wavelengths will be

`(m_(p)/m_(e))^(1//2)`

`m_(e)//m_(p)`

`m_(p)//m_(e)`

An electron of charge e and mass m is accelerated from rest by a potential difference V . the de Broglie wavelength is

Directly proportional to the square root of potential difference.

Inversely proportional to the square root of potential difference.

Directly proportional to the square root of electron mass

Inversely proportional of the cube root of electron mass