Dual Nature of Radiation and Matter MCQ

Ans: (a) dual nature

Explanation: The de Broglie equation is used to describe the wave properties of the matter. It states that matter can act as waves just as light and radiation also behave like waves and particles. Basically, it describes the dual nature of the matter. The waves that are associated with the particles that are moving are known as de Broglie waves or matter waves.

Ans: (b) 1.23 Å

Explanation: The value of V is given 100 volts.

Equation of de Broglie wavelength of electron λ=h2meV

\(\lambda=\frac{6.6 ×10^{-34}}{2(9.1×10^{-31})(1.6×10^{-19})(100)}\)

λ = 6.6 ×10^-34 / 5.4 ×10^-10m

\(\lambda=\frac{6.6 ×10^{-34}}{5.4×10^{-10}}m\)

\(\lambda=1.277 ×10^{10}m = 1.23 Å \)

Ans: (a) 1:2

Explanation: the specific charge = \(\frac{charge}{mass}\) ratio. A particle is a helium atom with a 2 unit positive charge having mass four times that of the proton. 

Let the charge on the proton be p and mass on the proton be m.

So, the specific charge of the proton is given by, s = p/m

Now, the charge on the proton is 2p and the mass is 4m.

Therefore, a specific charge \(S' = \frac{2p}{4m} = \frac{1p}{2m} = \frac{s}{2}\)

⇒\(\frac{s'}{s}=\frac{1}{2}\)

Ans: (b) E^1/2

Explanation: for proton, \(E=\frac{1}{2}mv^2=\frac{1}{2}\frac{m^2v^2}{m}\)

\(mv=\sqrt{2mE}\) Or \(\lambda_1=\frac{hc}{mv}=\frac{h}{\sqrt{2mE}}\)

For photon, \(E=hv_2=\frac{hc}{\lambda_2}\) or \(\lambda_2=\frac{hc}{E}\)

Therefore, \(\frac{\lambda_1}{\lambda_2}=\frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}\propto\frac{E}{\sqrt{E}}=E^{1/2}\)

Ans: (d) All of these

Explanation: Einstein’s photoelectric equation is given by,
\(hv-hv_0=\frac{1}{2}mV_{max}^2\)  or  \(hv-hv_0=\frac{1}{2}mV_{max}^2=eV\)

From the equations above, we can say that the maximum velocity of the electron reacting with the collector depends on the frequency of the incident radiation (v), on the work function (hv₀) and the potential difference between the emitter and the collector(V).

Ans: (a) 8 × 10¹⁴ Hz

Explanation: In the given question work function \(\phi\) = 3.3 eV

Formula for the photoelectric effect is given by hf_c = \(\phi\)

Therefore, threshold frequency f_c = \(\frac{\phi}{h}\) = \(\frac{(3.3 eV)(1.6×10^{-19}J/eV)}{6.626×10^{-34}Js}\) = 8×10¹⁴Hz

Ans: (b) atomic number

Explanation: Moseley’s law is a law concerning the characteristics of x-rays emitted by atoms.

It states that the square root of the frequency of the emitted x-ray is proportional to the atomic number (\(\sqrt {v}Z\)) as shown in the figure, where Z is the atomic number of elements. Therefore, Mosley's law relates the X-rays with the atomic number of the element.

Ans: (c) Blue light

Explanation: Einstein’s photoelectric equation is given by E_k = hv – \(\phi\)

Where, E_k is the maximum kinetic energy of the ejected electron.

Energy of the photon of the blue light (hv)_blue is higher compared to red light (hv)_red. In photoelectric emission, E_k \(\propto\) (hv). Therefore, blue light emits electrons of greater kinetic energy than that of red light.

Ans: (d) 2hv/3

Explanation: By using Einstein’s photoelectric equation, we have

\(\frac{1}{2}mV^{2} = hv - \phi_0\)

When the velocity V and frequency (ν) gets doubled, we get

\(\frac{1}{2}m(2V)^{2} = h2v - \phi_0\)

\(4(\frac{1}{2}mV^{2}) = 2hv - \phi_0\)

\(4(hv - \phi_0) = hv - \phi_0\)

\(2hv=3\phi_{0}\)

Therefore, work function for this metal is

\(\phi_{0} = \frac{2hv}{3}\)

Ans: (c) λ/2<λ’< λ

Explanation: As given in the question, \(E_k=\frac{hc}{\lambda}-\phi_0\)

When kinetic energy is changed to 2E we have the equation \(2E_k=\frac{hc}{\lambda}-\phi_0\)

There are two relations that can derived from the given equation,

λ’ > λ/2 and λ’ < λ

Therefore, the correct answer is option (c) λ/2<λ’< λ

Ans: (a) K₁< (K₂/3)

Explanation: \(K_{1} = \frac{hc}{\lambda_{1}} - \phi_0\)................. (I)

\(K_{2} = \frac{hc}{\lambda_{2}} - \phi_0\)

\(\frac{hc}{\lambda_{2}} = K_2 + \phi_0\)................... (II)

\(\therefore K_1-K_2 = hc[\frac{1}{\lambda_1} - \frac{1}{\lambda_2}]\)

\(= hc[\frac{1}{3\lambda_1} - \frac{1}{\lambda_2}] = - \frac{2hc}{3\lambda_2}\)

Now, by using equation (II), we get

\(K_1 - K_2 = - \frac{2}{3}(K_2 + \phi _0)\)

\(K_1 = K_2 - \frac{2}{3}K_2 - \frac{2}{3} \phi _0 = \frac{K_2}{3} - \frac{2}{3} \phi_0\)

\(\therefore K_1 < \frac {K_2}{3}\)

Ans: (d) \(\sqrt{\frac{m_e}{2m_a}}\)

Explanation: By using a photoelectric equation we can write \(\frac{1}{2}mv^2=eV\)

⇒\(mv = \sqrt{2meV}\)

Therefore, the ratio of momentum of an electron and an alpha particle is given by

\(\frac{\text{Momentum of electron}} {\text{Momentum of alpha particle}} = \frac{P_e}{P_a} = \sqrt{\frac{2m_e(e)V}{m_a(2e)V}} = \sqrt{\frac{m_e}{2m_a}}\)

Ans: (a) increasing the potential difference between the anode and filament

Explanation: if, e is the charge on electron, V is the potential difference, m is the mass of the particle and v is the velocity of the particle then, eV = ½ mv²

Therefore, in the Davisson and Germer experiment, velocity of electrons emitted from electron guns can be increased by increasing the potential difference between anode and filament.

Ans: (b) remains same

Explanation: Incident radiation is said to be radiation falling on an object and the intensity of incident radiation is not dependent on stopping potential, work function and maximum kinetic energy. Stopping potential depends on the frequency, wavelength of the incident light and material of the metal. Thus, the stopping potential will remain the same even if the incident radiation in a photocell is increased.

Ans: (a) 0

Explanation:

When the frequency of incident radiation becomes equal to threshold frequency, the value of stopping potential becomes zero. 

By Einstein’s photoelectric equation

hv = hv₀ + eV₀

As given in the question v = v₀

 Hv₀ = hv₀ + eV₀

v₀ = 0

Therefore, when the frequency of incident radiation becomes equal to threshold frequency, the value of stopping potential becomes zero.