Question

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The wavelength associated with a moving electron depends on it's mass(m) it's velocity (v) and Plank's constant (h).Derrive the equation for wavelength using the method of diamensions????

Answer 1

Wavelength dimension=[L] 
mass dimension=[M] 
velocity dimension= [LT^-1] 
plank constant dimension=[ML^2T^-1] 
now  
[L]=k [M]^a [LT^-1] ^b [ML^2T^-1] ^c 
=k [M]^(a+c)[L]^(b+2c)[T]^(-b-c) 
compare LHS to RHS 
a+c=0 
b+2c=1 
-b-c=0 
so  
c=1 
b=-1 
a=-1

hence 

lemda=k .plank constant/(mass. velocity)

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Answer 2

Let wavelength = l

Mass = m

Velocity = v

Plank's constant = h

$l \: \: \alpha \: \: {(m)}^{a} \: \: \: \: - - - (1) \\ l \: \: \alpha \: \: {(v)}^{b} \: \: \: \: - - - (2)\\ l \: \: \alpha \: \: ( {h)}^{c} \: \: \: \: - - - (3)$

$l \: \: \alpha \: \: {(m)}^{a} {(v)}^{b} ( {h)}^{c} \\ l = k{(m)}^{a} {(v)}^{b} ( {h)}^{c} \: \: \: \: \: \: \: \: - - - (4)$

${m}^{0} {l}^{1} {t}^{0} = k( { {m}^{1} })^{a} ( { {l}^{1} {t}^{ - 1}) }^{b} ( { {m}^{1} {l}^{2} {t}^{ - 1} })^{c} \\ {m}^{0} {l}^{1} {t}^{0} = k( {m}^{a + c} \: {l}^{b + 2c} \: {t}^{ - b - c} )$

$a + c = 0 \\ a + 1 = 0 \\ a = - 1\\ \\ b + 2c = 1 \\ b = 1 - 2c \\ b = 1 - 2 \\ b = - 1 \\ \\ - b - c = 0 \\ b = - c \\ 1 - 2c = - c \\ c = 1$

From (4)

$l = k( {m}^{ - 1} )( {v}^{ - 1} )( {h}^{1} ) \\ \\ l = \frac{kh}{mv} \\ \\ k \: \: \alpha \: \: \frac{h}{mv}$

Please take all the dimensional formulas in capital letters.

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