de-Broglie wave length of Bohr's first orbit is `lambda` .What is the circumference of third orbit? (B) 3 `lambda` B)9 `lambda` (C) 6 `lambda` (D) 4`lambda`

If the de Broglie wave length of the fourth Bohr's orbit of hydrogen atom is 4 Å, the circumference of the orbit is:

De Broglie wavelength lambda is proportional to

de Broglie wavelength of a moving particle is lambda . Its momentum is given by :

An electron is moving in Bohr's fourth orbit. Its de Broglie wavelength is lambda . What is the circumference of the fourth orbit?

The de Broglie wavelength of an electron moving in a circular orbit is lambda . The minimum radius of orbit is :

de-Broglie wavelength of an electron in the nth Bohr orbit is lambda_(n) and the angular momentum is J_(n) then

The intensity of X-rays form a Coolidge tube is plotted against wavelength lambda as shown in the figure. The minimum wavelength found is lambda_c and the wavelength of the K_(alpha) line is lambda_k . As the accelerating voltage is increased (a) lambda_k - lambda_c increases (b) lambda_k - lambda_c decreases (c ) lambda_k increases (d) lambda_k decreases

Electrons with de - Brogli wavelengtyh lambda fall on the target in an X-ray tube.The cut off wavelength of the emitted Xrays is (a) lambda_0 = (2mclambda^2)/(h) (b) lambda_0 = (2h)/(mc) (c ) lambda_0 (2m^2 c^2 lambda^3)/(h^2) (d) lambda_0 = lambda

The de-broglie wavelength of neutron at 27^(@)C is lambda . The wavelength at 927^(@)C will be

If vec a is a nonzero vector of magnitude a and lambda a nonzero scalar, then lambda vec a is unit vector if (A) lambda = 1 (B) lambda= -1 (C) a = |lambda| (D) a = 1/(|lambda|)