Question

wavelength of matter wave may depend upon plank constant ( h ) , mass (m) , velocity (v) of the particle .Now derived the formula .

Answer 1

It is the de broglie wavelength derivation
E=mc^2
E=h v( v =frequency)
then combining these two equations
mc^2=hv
real particles can't travel with speed of light so replace c with v(v=velocity)
mv^2=hv
subsitute v/lambda(v=velocity) for v(frequency)
mv^2=hv/lambda
lambda=hv/ mv^2
lambda=h/mv

Answer 2

Answer:

Concept:

A periodic wave's spatial period, or the distance across which the wave's structure repeats, is its wavelength. It's the separation between two adjacent matching places on the wave of the same phase, such as two adjoining emblems, troughs, or zero crosses, and it's a characteristic of both traveling and resting waves, and other spatially wave patterns. The opposite of the wavelength is the spatial frequency. The Greek letter lambda () is widely used to represent wavelength. The term wavelength is also used to describe modulated waves, as well as their sinusoidal envelopes or waves created by the interference of numerous sinusoids.

Given:

The wavelength of a matter wave may be affected by the particle's plank constant (h), mass (m), and velocity (v).

Find:

Now derived the formula.

Answer:

The peak is the highest point on a wave, whereas the valley is the lowest. The wavelength can be calculated using the formula below. wavelength = wave velocity/frequency. The length of a wave is usually measured in metres. The Greek lambda is the sign for wavelength, therefore = v/f.

λ $=kh^{a}$

  $=m^{b}v^{c}$

$[M^{0}L^{1}T^{00} ]=[ML^{2}L^{-1} ]^{a} M^{b} [LT^{-1}]^{c}$

                 $=M^{a+b} L^{2a+c} T^{-a-c}$

Using the idea of dimension homogeneity, we get at

$a+b=0 = > (1)$

$2a+c=1 = > (2)$

$-a-c=0 = > (3)$

$add (2) and (3)$

$we$ $get$

$a=1$

$from$  $(1)$ $b=-a=-1$

$from$ $(3)$  $c=-a=-1$

λ $=kh/$µ

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