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
$h / m u$
$\infty$
$h / m u s i n θ$
$h / m u c o s θ$

Question

A particle of mass m is projected from ground with velocity u making angle theta with the vertical- The de Broglie wavelength of the particle the highest point ish-muinftyh-mu sinthetah-mu costheta

Toppr Admin · Accepted Answer

De Broglie wavelength is given by-xA0-x3BB-hmvAt the highest point- only the horizontal component of the velocity will be there- which is usin-x3B8-Thus- wavelength is given by- -x3BB-hmusin-x3B8-

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 ish/mu∞h/musinθh/mucosθ

De 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
$h / m u$
$\infty$
$h / m u s i n θ$
$h / m u c o s θ$

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, $λ = \frac{h}{m u sin θ}$