Question

SIRF JISKO AATA HO WAHI KAREE SOLVE.....✔✔

Answer 1

║║║║║║hello mate║║║║║║║║

АпЗЩЁГ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓

the answer is ( c ) i belive , cause u put a ∨¬ for any electronic equation

АпЗЩЁГ↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑

plz mark me brainliest if u liked it

# be brainly ФΔπω⊅⊄⊇

Answer 2

$\underline{\underline{\colorbox{cyan}{HeY Mate!!}}}$

➡we can use De-Broglie wavelength and kinetic energy formula that is in electron volt

De-Broglie wavelength for an electron(lambda) = h/p

p = h / lambda............[1]

Kinetic energy of one 1 electron = eV

So , we also know that the kinetic energy = P²/2m

eV = P²/2m

=>
$p = \sqrt{2mev}$
..........[2]

so by putting the value from (1) in (2) we get

h/lambda =
$\sqrt{2mev}$
So your answer is (c) is the required answer

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$\textcolor{white}{{\boxed{\colorbox{black}{Hope it helps BE BRAINLY}}}}$