The threshold energy is given as E0 and radiation of energy E falls on metal, then K.E. is given as

Question

The threshold energy is given as E0 and radiation of energy E falls on metal, then K.E. is given as (A) ( E - E 0/2) (B) E - E 0 (C) E 0- E (D) (E/E0)

Tardigrade Admin · Accepted Answer

E = Incident energy E 0= Threshold energy E= E 0+ K . E K.E.= E - E 0

Q. The threshold energy is given as $E_{0}$ and radiation of energy $E$ falls on metal, then K.E. is given as

$\frac{E - E _{0}}{2}$

$E - E_{0}$

$E_{0} - E$

$\frac{E}{E _{0}}$

$E =$ Incident energy
$E_{0} =$ Threshold energy
$E = E_{0} + K . E$
$K . E . = E - E_{0}$

Q. The threshold energy is given as E0​ and radiation of energy E falls on metal, then K.E. is given as

2E−E0​​

E−E0​

E0​−E

E0​E​