The De Broglie Explanation

Q. A photon and an electron have equal energy E. Then the ratio of wavelength of photon to that of electron is proportional to

1. $\sqrt{E}$
2. $\frac{1}{\sqrt{E}}$
3. $E$
4. $\frac{1}{E}$

Q. What is the de Broglie wavelength of an electron with a kinetic energy of $120\text{eV}$?
(Given: $h=6.63\times {10}^{-34}\text{Js},m_e=9.11\times {10}^{-31}\text{kg},e=1.6\times {10}^{-19}\text{coulomb}$

1. $725\text{pm}$
2. $500\text{pm}$
3. $322\text{pm}$
4. $112\text{pm}$

Q.

According to De Broglie 2$\pi$r=n$\lambda$ where n is the quantum number of an orbit, in visual representation of an electron as a wave, n stands for:

1. No. of loops , where one loop is one wavelength

2. No. of electrons

3. No. of half wavelengths

4. Can not comment

Q. Which of the following has the largest de Broglie wavelength assuming that the velocity is the same for all?

1. $C{O}_2$ molecule
2. $N{H}_3$ molecule
3. electron
4. proton

Q. The de-Broglie wavelength of an electron in first orbit of Bohr's hydrogen is equal to

1. radius of the orbit
2. perimeter of the orbit
3. diameter of the orbit
4. half of the perimeter of the orbit

Q. One of the lines in the emission spectrum of ${\text{Li}}^{2+}$ has the same wavelength as that of $2^{nd}$ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is :-

1. $n=12\to n=6$
2. $n=4\to n=2$
3. $n=8\to n=2$
4. $n=8\to n=4$

Q. According to De Broglie the electron in n=3 of Hydrogen atom will look like

1. 
2. 
3. 
4. 

Q. The circumference of first orbit of hydrogen atom is ‘s’. Then the Broglie wavelength of electron in that orbit is

1. $\frac{S}{2}$
2. 2S
3. S
4. 3S

Q.

The de-Broglie wavelength of an electron in the first Bohr orbit is

1. Equal to one fourth the circumference of the first orbit

2. Equal to half the circumference of the first orbit

3. Equal to twice the circumference of the first orbit

4. Equal to the circumference of the first orbit

Q. The de-Broglie wavelength of an electron in first orbit of Bohr's hydrogen is equal to

1. radius of the orbit
2. perimeter of the orbit
3. diameter of the orbit
4. half of the perimeter of the orbit

Q. According to De Broglie the electron in n=3 of $L{i}^{++}$ atom will look like

1. 
2. 
3. 
4.