If the potential difference applied across X- | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

${\lambda _{\min }} = \frac{{hc}}{{eV}} = \frac{{6.6{ 	imes ^{ - 34}} 	imes 3 	imes {{10}^8}}}{{1.6 	imes {{10}^{ - 19}}V}}$
$ = \frac{{12375}}{V

Vedclass · Accepted Answer

$\frac{{12400}}{V}\;\mathring A$

Vedclass · Answer

${\lambda _{\min }} = \frac{{hc}}{{eV}} = \frac{{6.6{ 	imes ^{ - 34}} 	imes 3 	imes {{10}^8}}}{{1.6 	imes {{10}^{ - 19}}V}}$
$ = \frac{{12375}}{V} \approx \frac{{12400}}{V}\;\mathring A$

If the potential difference applied across $X-$ ray tube is $V$ volts, then approximately minimum wavelength of the emitted $X-$ rays will be

Similar Questions

Explain Rutherford's argument for scattered $\alpha $ -particles.

According to classical theory, the circular path of an electron in Rutherford atom is

The energy of hydrogen atom in $n^{th}$ orbit is $E_n$, then the energy in $n^{th}$ orbit of singly ionised helium atom will be

According to the nuclear model of an atom, what is the area of the whole mass of the atom located ?

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion