A stream of electrons from a heated filaments was passed two charged plates kept at a potential difference $$V$$ $$esu.$$ If e and m are charge and mass of an electron, respectively, then the value of $$\frac{h}{\lambda }$$ ( where $$\lambda $$ is wavelength associated with electron wave ) is given by :
A.
$$\sqrt {meV} $$
B.
$$\sqrt {2meV} $$
C.
$$meV$$
D.
$$2meV$$
Answer :
$$\sqrt {2meV} $$
Solution :
As electron of charge $$'e’$$ is passed through $$'V'$$ volt, kinetic energy of electron will be $$eV$$
Wavelength of electron wave $$\left( \lambda \right) = \frac{h}{{\sqrt {2m.K.E} }}$$
$$\lambda = \frac{h}{{\sqrt {2meV} }} \Rightarrow \,\frac{h}{\lambda } = \sqrt {2meV} $$
Releted MCQ Question on Physical Chemistry >> Atomic Structure
Releted Question 1
The number of neutrons in dipositive zinc ion with mass number 70 is