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This is second of our many quizzes on Modern physics. Click on this to access more such competitive quizzes on Modern Physics and Semiconductors.

Modern physics Quiz 2

physics

The heat of combustion of gasoline is $1.13\times 10^4\,kcal\,kg^{-1}$. If the average binding energy per nucleon is $8\,MeV$, how many times larger is the binding energy in comparison with the combustion heat of gasoline? $(\text{take mass of a nucleon} = 1.66 \times 10^{-27}\,kg, 1 MeV = 1.6\times 10^{-13}\,J)$ 

Find the  $BE$ per nucleon using the given nucleon mass. Convert this  $BE$ to kcal and compare the powers of the  $BE$ of the nucleus and the heat of combustion of gasoline.  

  $\text{ We are given that the binding energy per nucleon is 8 MeV, i.e.,}\\ BE=  \dfrac{8 MeV}{nucleon}\\ \text{ And the mass of the nucleon}\, m = 1.66 \times 10^{-27} kg\\ BE = \dfrac{8 \times 1.6 \times 10^{-13}}{1.66 \times 10^{-27}}\\ \Rightarrow BE = 7.71 \times 10^{14} Jkg^{-1}\\ \text{ Converting our  BE from}\, J kg^{-1}\,\text{to}\, kcal kg^{-1}\,\text{we get:}\\ BE = \dfrac{7.71\times 10^{14}}{4.2 \times 10^{3}} \\ \Rightarrow BE= 1.84 \times 10^{11} kcal kg^{-1}\\ \text{ Given that the heat of combustion of gasoline is}\, 1.13 \times 10^{4} kcal kg^{-1}\\ \text{We see that the BE is}\, \approx 10^{11-4} = 10^{7}\,\text{times greater.}$

What is the minimum amount of energy that an electron should possess in order to be able to exist inside the nucleus of an atom, whose nuclear diameter is of the order $10^{-14}\,m$? $(\text{Take h} = 6.626 \times 10^{-34}\,Js)$ 

The de Broglie wavelength of such an electron will be limited to the diameter of the nucleus.  

  $\text{ If an electron is to exist inside a nucleus,}\\ \text{its de Broglie wavelength must not be larger than the nuclear diameter,}\\ \text{which is of the order}\, 10^{-14} m.\\ \text{The momentum possessed by such an electron can be given by:}\\ p = \dfrac{h}{\lambda} \\ \Rightarrow p= \dfrac{6.626 \times 10^{-34}}{10^{-14}} \\ \Rightarrow p= 6.626 \times 10^{-20} Jm^{-1}s\\ \text{The energy possessed by such an electron will be:}\\ E = pc \\ \Rightarrow E = 6.626 \times 10^{-20} \times 3 \times 10^{8} \\ \Rightarrow E= 19.89 \times 10^{-12}J\\ \text{Converting it into MeV:}\\ E = \dfrac{19.89 \times 10^{-12}}{1.6 \times 10^{-13}} \\ \therefore E= 124.3 \approx 124\,MeV$

The following graph shows the activity of a certain radioactive sample plotted against time. If initially the slope of the curve is $m$, then determine the slope of the curve at point $P$ which is situated at the mean-life mark as  shown in the figure.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a2e82079-387b-4b64-a172-75cad646a59c919287475316260589.png"/> 

Use the equation of radio-activity to arrive at the general slope expression, and plug in values specific to each case to arrive at appropriate slope expression for $P$.

 $\text{We are given that when t=0, the slope of the graph is}\,\dfrac{dA}{dT} = m.\\ \text{From the equation of activity}\, A= A_0e^{-\lambda t}.\\ \text{The slope of the graph at any point is:}\\\dfrac{dA}{dt} = \dfrac{d}{dt}( A_0e^{-\lambda t})\\  \Rightarrow\dfrac{dA}{dt}  = -\lambda A_0e^{-\lambda t}\\ \text{ The slope of the graph at t=0 will be:}\\ \dfrac{dA}{dt} = -\lambda A_0\\ \Rightarrow -\lambda A_0 = m\\ \text{Therefore, the slope at point P where}\, t=\tau\,\text{will be:}\\ \left(\dfrac{dA}{dt}\right)_P = -\lambda A_0e^{-\lambda \tau} \\ \Rightarrow \left(\dfrac{dA}{dt}\right)_P = me^{-\lambda \tau}\\ \text{We know that,}\\ \tau = \dfrac{1}{\lambda} \\ \Rightarrow \lambda \tau = 1 \\ \Rightarrow \left(\dfrac{dA}{dt}\right)_P = me^{-1} \\ \therefore\left(\dfrac{dA}{dt}\right)_P =\dfrac{m}{e}$

A charged oil drop falls under gravity with a terminal speed $v$. By applying a suitable electric field, the drop is held stationary and is found to carry an excess of $4$ electrons, following which the drop suddenly moves upwards with the speed $v$. Determine the total excess number of electrons possessed by the oil drop after moving upwards. 

When the drop falls under gravity, its weight is balanced by the viscous drag. When it is held stationary via an electric field, the electric force acting on the oil drop balances out the weight of the drop. Here, it gains electrons following which it experiences an electric force greater than the weight of the drop which makes it move upwards.  

 $\text{When the drop falls downwards under the influence of gravity,}\\ \text{at terminal velocity v, the apparent weight of the drop will be equal to}\\ \text{the viscous drag acting on the drop, i.e.,}\\ W = 6\pi\eta rv\\ \text{Next, the drop is held stationary by applying an electric field E:}\\ W = F_{electric} = Eq = E(4e)\\ \text{The drop  now moves upwards with the same velocity v with an electric force acting on it:}\\ W = F^{\prime} + 6\pi\eta rv\\ \Rightarrow F^{\prime} = W + 6\pi\eta r v\\ \Rightarrow Eq^{\prime} = W +6\pi\eta r v\\ \text{Let the number of electrons picked up by the drop from the applied field be n.}\\ E(ne) = W+ 6\pi\eta rv\\ \text{But previously, we found that}\, W=6\pi\eta rv\\ E(ne) = W+W =2W\\ \text{And we also found that}\,W = E(4e)\\ \Rightarrow E (ne) = 2 \times E(4e) \\ \Rightarrow n = 4\\ \text{ The drop thus has 4+4=8 excess electrons in total.}$

Determine the decay rate and the half-life period using the following graph.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0a15ce8a-0dea-4a2c-8439-109260b0ff7a7624259403950360285.png"/> 

Start with the exponential decay expression, take logarithm on both sides to arrive at an equation with slope of the graph which is equivalent to $\theta$. Determine the decay constant and plug it into the half-life equation. 

 $\text{We know that}\, N = N_0 e^{-\lambda t}\\ \text{Taking logarithm on both sides and rearranging:}\\ log N = \left(\dfrac{-\lambda}{2.303}\right)t +log N_0\\ \text{The above expression is similar to y =mx+c.}\\ \text{Equating the slope of the equation and slope from the graph expressions:}\\ \dfrac{-\lambda}{2.303} = -tan \theta\\ \Rightarrow \lambda = 2.303tan \theta\\ \text{Now, we know that the half life period is given by the expression:}\\ t_{1/2} = \dfrac{2.303}{\lambda}log_{10}(2)\\ \text{Plugging in the decay constant we get:}\\ t_{1/2} = \dfrac{2.303log_{10}(2)}{2.303tan \theta}\\ \Rightarrow t_{1/2} = log_{10}(2).\dfrac{1}{tan \theta} \\ \therefore t_{1/2} = log_{10}(2)cot \theta$