Saved Bookmarks
| 1. |
For electron moving with speed of 20 m/s and accelerated by 120 V p.D. de-Broglie wavelength can be easily obtained. |
|
Answer» Solution :Let stationary electron having mass m and CHARGE e is accelerated by voltage V. Kinetic energy of electron =voltage (p.d.) K=eV But `K=(1)/(2)mv^(2)` Multiplying and DIVIDING by m, `=(1)/(2)(m^(2)v^(2))/(m)` p=mV `p^(2)=m^(2)V^(2)` `THEREFORE K=(p^(2))/(2m)` `therefore p^(2)=2mK` `therefore p=SQRT(2mK)` k=eV `therefore p=sqrt(2meV)` `therefore lambda=(h)/(p)=(h)/(sqrt(2meV))` `therefore lambda=(h)/(sqrt(2meV))` Substituting value `m=9.11xx10^(-31)`,e`=1.6xx10^(-19)`, V=120 V, `h=6.625xx10^(-34)` `lambda=(6.625xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx120))` 0.112 nm |
|