During the transmission of nerve impulse through a nerve fibre, the potentila on the inner side of the plasma membrane has which type of electric charge?

A beam of charged particles of charge q and a mass m are accelerated from rest through a potential difference of 100V. They pass through crossed electric and magnetic fileds which together produce null deflection .If these electronic and magnetic fields are respectively 15xx10^(3)Vm^(-1)and 5Wbm^(-2) ,then q/m has a value, in C kg^(-1) ,equal to

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The orbital angular momentum of an electron in p-orbital makes an angle of 45^@ from Z-axis. Hence Z-component of orbital angular momentum of election is :

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) he permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The maximum orbital angular momentum of an electron with n= 5 is

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition mush compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The spin-only magnetic moment of free ion is sqrt(8) B.M. The spin angular momentum of electron will be

Electrolysis is the process in which electrical energy is converted to chemical energy. In electrolyte cell, oxidation takes place at anode and reduction at cathode. Electrode process depends on the electrode taken for electrolysis. Amount of substance liberated at an electrode is directly proportionation to the amount of charge passed through it. The mass of substance liberated at electrode is calculate using the following realation : m=(itE)/(96500) Here, E represent the equivalent mass and 96500 C is called the faraday constant. Faraday (96500 C) is the charge of 1 mole electron i.e., 6.023 xx 10^(23) electrons, it is used to liberate on gram equivalent of the substance. Calculate the volume of gas liberated at the anode at STP during the electrolysis of a CuSO_4 solution by a current of 1 A passed for 16 minutes and 5 seconds.

Fill in the blanks with suitable words given below. Fluctuation s of hormone (i)_____________ levels results in sensation of hunger and motivation of consuming food. When you feel your stomach is full and there is no need of food any more. Another hormone (ii)_____________ that gets secreted suppresses hunger. When we take food into the mouth it has to be chewed thoroughly. For this purpose the (iii)_____________ muscles help in chewing actions, while the (iv)_____________ muscles of the jaw moves the jaw up,down ,forward and backward during food mastication. The (v)_____________ nerve controls the muscles of the jaw. . Under the action of (vi)_____________ nervous system Saliva is released by the salivary glands moistens the food to make chewing and swallowing easier. The salivary (vii)_____________ in the saliva breaks down the starch into sugars. As a result of chewing the food is transported into the oesophagus by the action of swallowing which is coordinated by the swallowing centre in the (viii)_____________ and the (ix)_____________. The tongue which is gustatory recognizes the taste and (x)_____________ nerve plays an important role in sensation of taste. Choose the right ones. 1) leptin, grehlin gastrin secretin. 2) ghrelin leptin secretin gastrin. 3) deep muscles ,surface muscles ,circular muscles, striated muscles. 4) surface muscles, deep muscles, neck muscles, long muscle. 5) fifth cranial nerve ,second cranial nerve,fifth facial nerve, spinal nerve. 6) central nervous system, peripheral nervous system autonomous nervous system. 7) lipase, sucrase, galactase,amylase. 8) medulla oblongata, cerebrum , 8th spinal nerve,cranial nerve.7th cranial nerve. 9) Pons varoli, brain stem ,medulla oblongata, mid brain. 10) 6^(th) cranial nerve, 5^(th) cranial nerve, 10^(th) cranial nerve, optic nerve.