Question

The de Broglie wavelength of a 66 kg man sking down
Kufri Hill in Shimla at 1x10 m sec- is:
(a) 1x10-36 m
(b)1x10-37 m
(c)1x10-38 m
(d)1x10-39 m​

Answer 1

Answer:

Hey mate ,

Your answer will be

(a) 1* 10^-36

Explanation:

Lambda = h/mv

lambda= ?  

h = 6.63 * 10^-34

m= 66kg

v = 1 *10 m/s

Lambda =  6.63 * 10^-34/ 66*1*10

lambda= 1*10^-36 m

Hope this will help uh

Answer 2

Answer:

OPTION (c) $1 \times 10^{-38}$ m

Explanation:

We know that, according to the de Broglie wavelength formula, λ = $\frac{h}{mv}$

where λ is the wavelength,

$h$ is the Planck constant,

$m$ is the mass, and

$v$ is the velocity.

Substituting the values in the above formula we get,

λ = $\frac{6.62 \times 10^{-34} }{66 \times 10}$

λ = $\frac{0.100 \times 10^{-34} }{10}$

λ = $1 \times 10^{-38}$ m

Hence, the de Broglie wavelength of a $66$ kg man skiing down Kufri Hill in Shimla at $1 \times 10 \ m/s$ is $1 \times 10^{-38}$ m.

#SPJ3