Question

solve fastt...
derive an expression for wavelength(lamda) which depends upon kinetic energy and mass..

Answer 1

Hope u like my process
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We know,

$= > \bf \: frequency (v)= \frac{velocity \: \: of \: \: light \: (c)}{wavelength( \lambda)}$
Also..

We know that,

$= > \bf \: E = m {c}^{2} ......( \underline{einstein's \: \: equation)} \\ \\ = > \bf{E} = hv \\ \\ where \: \: v = frequency \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: h = planck's \: \: constant$
So, we can say that,

$= > \bf \: m {c}^{2} = hv \\ \\ or. \: \: m {c}^{2} = \frac{hc}{ \lambda} \\ \\ or. \: \bf\: \lambda = \frac{hc}{m {c}^{2} } = \frac{hc}{K.E.} \\ \\ \\ = > \bf \lambda = \frac{h}{mc} = \frac{hc}{K.E.}$
Thus

The relation of wavelength with mass(m) and kinetic energy(K.E.) is derived.

Now we all know that

=> K.E. = ½ mv²

And also derive lambda = h/(mv)

So, can say that mv = h/(lambda)

So,

=> 2 × K.E. = mv²

=> (2×K.E.)×m= (mv)²

=>${ (\frac{h} { \lambda}) }^2$ = 2m×K.E.

=> $(\lambda)^2$ = $\frac{h^2}{ 2m \times K.E.}$

=>wavelength( $\lambda$ ) =$\frac{h} {\sqrt{2m \times K.E. } }$

So hope this is what u required too
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