An \alpha -particle and a proton are accelerated from rest by a potential difference of 100 V . After this, their de Broglie wavelengths are \lambda_{\alpha} and \lambda_{p} respectively. The ratio \dfrac{\lambda_{p}}{\lambda_{\alpha}} , to the nearest integer, is:

We have, \dfrac{1}{2}mv^{2}=eV De broglie wavelength= \lambda=\dfrac{h}{mv} \lambda = \dfrac{h}{\sqrt{2meV}} \lambda_{p}=\dfrac{h}{\sqrt{2(m)eV}} ....(1) \lambda_{\alpha}=\dfrac{h}{\sqrt{2(4m)2eV}} ....(2)Now, \dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{8} \simeq 3

Dual Nature Of Radiation And Matter - Types of questions

Q: A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is : (take the proton mass, m_p = (5/3) \times 10^{-27}kg and h/e = 4.2\times 10^{-15}J.s/C; \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 m/F; 1 fm = 10^{-15}m )

Given here, q_{1} = 120 \times e q_{2} = e ( charge on proton) It is based on conservation of energy of proton. \dfrac{1}{2}mv^2 = k \dfrac{q_1q_2}{R} v = \sqrt{\dfrac{2Kq_1q_2}{mR}} Also, we know, de’broglie wavelength, \lambda =\dfrac{h}{mv} \lambda = \dfrac{h\sqrt{R}}{\sqrt{2mKq_1q_2}}\\ Thus, \lambda = \dfrac{h\sqrt{10\times10^{-15}}}{\sqrt{2\times\dfrac{5}{3}\times10^{-27}\times9\times10^9\times120\times e^2}}\\ \lambda =\dfrac{\dfrac{h}{e}\times10^{-7}}{6\times10^{-8}}\\ \lambda =\dfrac{4.2\times10^{-15}\times10^{-7}}{6\times10^{-8}}\\ \lambda =7\times10^{-15}\ m = 7\ fm \\ Option A is correct.

Q: A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3\times 10^{8}m/s . The final momentum of the object is:

Sol . (B) t=100\times 10^{-9} sec,\ P=30\times 10^{-3} Watt, C=3\times 10^{8}m/s Incident energy = Pt By debroglie wavelength, \lambda=h/p and Energy of photon is E=hc/\lambda From above two equations, E=pc Momentum p=\dfrac{Pt}{C}=\dfrac{30\times 10^{-3}\times 100\times 10^{-9}}{3\times 10^{8}}=1.0\times 10^{-17} kg ms ^{-1} So, the answer is option (B).

Q: The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is

Slope of graph is h/e = constant \Rightarrow 1

Q: A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm . The maximum speeds of the photoelectrons corresponding to these wavelengths are u_1 and u_2 , respectively. If the ratio u_1 : u_2 = 2 : 1 and hc = 1240 eV nm , the work function of the metal is nearly :

On solving we get \displaystyle \dfrac{\dfrac{hc}{\lambda_1}- \phi}{\dfrac{hc}{\lambda_2} - \phi} = \dfrac{u_1^2}{u_2^1} On putting the values we get \phi = 3.7 eV

Q: The graph between 1/\lambda and stopping potential (V) of three metals having work functions \phi_{1},\ \phi_{2} and \phi_{3} in an experiment of photo-electric effect is plotted as shown in the figure. Which of the following statement(s) \mathrm{i}\mathrm{s}/\mathrm{a}\mathrm{r}\mathrm{e} correct? [Here \lambda is the wavelength of the incident ray].

From photo electric effect, eV=\dfrac{hc}{\lambda}-\phi where V= stopping potential and \phi= work function. Slope , m=\tan\theta=hc/e (compare above equation with y=mx-c )If V=0 \Rightarrow \dfrac{\phi}{hc}=\dfrac{1}{\lambda} Now from given graph,for metal 1 : \dfrac{\phi_1}{hc}=0.001 for metal 2: \dfrac{\phi_2}{hc}=0.002 for metal 3: \dfrac{\phi_3}{hc}=0.004 Thus, \phi_1:\phi_2:\phi_3=1:2:4 Also, \phi=\dfrac{hc}{\lambda_0} where \lambda_0 be the threshold wavelength. For metal 2 : \lambda_{02}=1/0.002=500 nm For metal 3 : \lambda_{02}=1/0.004=250 nm As the wavelength of violet colour is around 400 nm so only metal 2 can eject violet colour light but not metal 3. Thus, options A and C is right.