1.

Two particles A and B are in motion. If the wavelength associated with the particle A is 5xx 10^(-8) m , calculate the wavelength of particleB, if its momentum is half of A.

Answer»

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SOLUTION :According to de-Broglie relation,
`lamda = h/p " or " p = h/lamda `
for particle A , `p_A= (h)/(lamda_A)`
Here, `p_A` and `lamda_A`are the momentum and wavelength of particle A.
for particle B `p_B = (h)/(lamda_B)`
Here` p_B` and `lamda_B` are the momentum and wavelength of particle B.
But `p_B = 1/2 p_A`
` THEREFORE (h)/(lamda_B) = 1/2 (h)/(lamda_A)`
`(lamda_A)/(lamda_B) = 1/2 ` or`lamda_B = 2lamda_A`
But `lamda_A = 5 xx 10^(-8) m`
`lamda_B = 2lamda_A = 2 xx 5 xx 10^(-8) m = 10 xx 10^(-8) m= 10^(-7) m`


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