A particle of mass m is projected from ground with velocity u making angle - with the vertical- The de Broglie wavelength of the particle at the highest point is

The correct option is C h/mu sinθ De Broglie wavelength is given by: #160; λ=hmv At the highest point, only the horizontal component of the velocity ...

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Standard XII

Physics

Matter Waves

A particle of...

Question

A particle of mass m is projected from ground with velocity u making angle $θ$ with the vertical. The de Broglie wavelength of the particle at the highest point is

\infty

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h / m u s i n θ

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h / m u c o s θ

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h / m u

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Solution

The correct option is C $h / m u s i n θ$
De Broglie wavelength is given by: $λ = \frac{h}{m v}$

At the highest point, only the horizontal component of the velocity will be there, which is $u sin θ$ .

Thus, wavelength is given by, $λ = \frac{h}{m u sin θ}$

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A particle of mass m is projected from ground with velocity u making angle θ with the vertical. The de Broglie wavelength of the particle at the highest point is

The correct option is C h/musinθDe Broglie wavelength is given by: λ=hmvAt the highest point, only the horizontal component of the velocity will be there, which is usinθ.Thus, wavelength is given by, λ=hmusinθ

A particle of mass m is projected from ground with velocity u making angle $θ$ with the vertical. The de Broglie wavelength of the particle at the highest point is

The correct option is C $h / m u s i n θ$
De Broglie wavelength is given by: $λ = \frac{h}{m v}$

At the highest point, only the horizontal component of the velocity will be there, which is $u sin θ$ .

Thus, wavelength is given by, $λ = \frac{h}{m u sin θ}$