HSC Science (Computer Science) 12th Standard Board Exam - Maharashtra State Board Question Bank Solutions for Physics

Calculate De Broglie's wavelength of the bullet moving with speed 90m/sec and having a mass of 5 gm. 

Explain De Broglie’s Hypothesis.

The radii of Bohr orbit are directly proportional to ______ 

According to Bohr's second postulate, the angular momentum of the electron is the integral multiple of `h/(2pi)`. The S.I unit of Plank constant h is the same as ______ 

For the hydrogen atom, the minimum excitation energy ( of n =2) is ______

The speed of electron having de Broglie wavelength of 10 ^-10 m is ______ 
(m_e = 9.1 × 10^-31 kg, h = 6.63 × 10^-34 J-s) 

If a_O is the Bohr radius and n is the principal quantum number then, state the relation for the radius of n^th orbit of the electron in terms of Bohr radius and principal quantum number. 

What is the energy of an electron in a hydrogen atom for n = ∞?  

The linear momentum of the particle is 6.63 kg m/s. Calculate the de Broglie wavelength.

Starting with 𝑟 = `(ε_0h^2n^2)/(pimZe^2),` Show that the speed of an electron in n^th orbit varies inversely to principal quantum number. 

State Bohr's second postulate for the atomic model. Express it in its mathematical form.  

State any two limitations of Bohr’s model for the hydrogen atoms. 

Using de Broglie’s hypothesis, obtain the mathematical form of Bohr’s second postulate.

Calculate the longest wavelength in the Paschen series.

(Given R_H =1.097 ×10⁷ m^-1)  

The angular momentum of an electron in the 3^rd Bohr orbit of a Hydrogen atom is 3.165 × 10^-34 kg m²/s. Calculate Plank’s constant h.    

Derive an expression for the radius of the n^th Bohr orbit for the hydrogen atom.

Calculate the wavelength for the first three lines in the Paschen series. 
(Given R_H =1.097 ×10⁷ m^-1)  

Calculate the shortest wavelength in the Paschen series if the longest wavelength in the Balmar series is 6563 A^o. 

State the postulates of Bohr’s atomic model. Hence show the energy of electrons varies inversely to the square of the principal quantum number. 

Obtain an expression for wavenumber, when an electron jumps from a higher energy orbit to a lower energy orbit. Hence show that the shortest wavelength for the Balmar series is 4/R_H.